The compound that will not be soluble in water is BaSO₄ (barium sulfate). Option B is correct.
BaSO₄ (barium sulfate) is the compound that is not soluble in water. Solubility in water depends on the nature of the compound and the interactions between its constituent ions and water molecules. In the case of BaSO₄, the strong electrostatic forces of attraction between the barium (Ba²⁺) and sulfate (SO₄²⁻) ions result in a very low solubility in water. The solubility rules dictate that most sulfates are soluble in water, but barium sulfate is an exception.
It forms a solid precipitate when barium ions and sulfate ions come into contact in an aqueous solution. On the other hand, K₂S (potassium sulfide), NaNO₃ (sodium nitrate), and LiOH (lithium hydroxide) are all soluble in water and will dissociate into their constituent ions when dissolved, resulting in a homogeneous solution. Option B is correct.
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According to the Standard Reduction Potential Table in your textbook, assign the anode and cathode for thecell pairs, B1-B2, A1-B2, AND A2-B2 described in the Procedure
For the B₁-B₂ cell pair, B₂ is the cathode, and B₁ is the anode. For the A₁-B₂ cell pair, B₂ is the cathode, and A₁ is the anode. For the A₂-B₂ cell pair, B₂ is the cathode, and A₂ is the anode.
B₁-B₂:
The half-reaction at B₁ is given as A⁺(aq) + 2e⁻ → A(s) with a reduction potential of -0.78 V.
The half-reaction at B₂ is given as B⁺(aq) + e⁻ → B(s) with a reduction potential of -0.45 V.
Since the reduction potential of B₂ (-0.45 V) is more positive than B₁ (-0.78 V), B2 will be the cathode, and B1 will be the anode.
A₁-B₂:
The half-reaction at A₁ is given as C⁺(aq) + 2e⁻ → C(s) with a reduction potential of -0.95 V.
The half-reaction at B₂ is the same as mentioned before B⁺(aq) + e⁻ → B(s) with a reduction potential of -0.45 V.
Since the reduction potential of B₂ (-0.45 V) is more positive than A₁ (-0.95 V), B₂ will be the cathode, and A₁ will be the anode.
A₂-B₂:
The half-reaction at A₂ is given as D⁺(aq) + 2e⁻ → D(s) with a reduction potential of -0.70 V.
The half-reaction at B₂ is the same as mentioned before B⁺(aq) + e⁻ → B(s) with a reduction potential of -0.45 V.
Since the reduction potential of B₂ (-0.45 V) is more positive than A₂ (-0.70 V), B₂ will be the cathode, and A₂ will be the anode.
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Calculate the mass in grams of 8.35 × 10²² molecules of CBr₄.
To calculate the mass in grams of 8.35 × 10²² molecules of CBr₄ (carbon tetrabromide), we need to use Avogadro's number to convert the given number of molecules to moles and then use the molar mass of CBr₄ to convert moles to grams.
The molar mass of CBr₄ can be calculated by adding up the atomic masses of carbon (C) and four bromine (Br) atoms. The atomic mass of carbon is approximately 12.01 g/mol, and the atomic mass of bromine is approximately 79.90 g/mol.
Molar mass of CBr₄ = (1 × 12.01 g/mol) + (4 × 79.90 g/mol) = 331.74 g/mol
To convert the number of molecules to moles, we divide the given number of molecules by Avogadro's number (6.022 × 10²³ molecules/mol):
Moles of CBr₄ = (8.35 × 10²² molecules) / (6.022 × 10²³ molecules/mol) = 0.138 mol
Finally, to find the mass in grams, we multiply the number of moles by the molar mass:
Mass of CBr₄ = (0.138 mol) × (331.74 g/mol) = 45.80 g
Therefore, the mass in grams of 8.35 × 10²² molecules of CBr₄ is approximately 45.80 grams.
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Can you think of something you do or a hobby you have that is physics
related? Explain your thinking as to how what you do is Physics related.
You can take physics as a hobby and also it is great that you like both maths and physics as well and also want to be an computer science engineer because physics and math
here is a more complex redox reaction involving the permanganate ion in acidic solution: 5no2− 6h 2mno4− → 5no3− 2mn2 3h2o
The balanced equation of the redox reaction is:5NO2−+ 6H+ + 2MnO4− → 5NO3− + 2Mn2+ + 3H2O.
Redox reaction involving the permanganate ion in acidic solution:5NO2−+ 6H+ + 2MnO4− → 5NO3− + 2Mn2+ + 3H2OThe balanced redox reaction that involves the permanganate ion in acidic solution is given by:5NO2−+ 6H+ + 2MnO4− → 5NO3− + 2Mn2+ + 3H2OFor balancing the given reaction in acidic solution, we follow the given steps:Step 1: Writing the unbalanced reaction in the ionic form:Step 2: Separating the half-reactions for oxidation and reduction:Oxidation half-reaction: 5NO2− → 5NO3−Reduction half-reaction: 2MnO4− → 2Mn2+Step 3: Balancing the number of atoms of elements in half-reactionsBalancing oxidation half-reaction:5NO2− → 5NO3− + 10H+ + 2e-Balancing reduction half-reaction: 16H+ + 2MnO4− → 2Mn2+ + 8H2O + 5e-Step 4: Balancing the number of electrons lost and gained in each half-reaction by multiplying by a coefficient.Oxidation half-reaction: 5NO2− → 5NO3− + 10H+ + 2e-Balancing reduction half-reaction: 16H+ + 2MnO4− → 2Mn2+ + 8H2O + 5e-Now, the electrons lost and gained are balanced.Step 5: Balancing the number of hydrogen ions in each half-reaction:Oxidation half-reaction: 5NO2− + 10H+ → 5NO3− + 2e-Balancing reduction half-reaction: 2MnO4− + 16H+ → 2Mn2+ + 8H2O + 5e-Step 6: Balancing the number of oxygen atoms in each half-reaction by adding H2O molecules to the side that needs it. For acidic solution, add H+ ions to balance the oxygen atoms.Oxidation half-reaction: 5NO2− + 10H+ → 5NO3− + 2e-Balancing reduction half-reaction: 2MnO4− + 16H+ → 2Mn2+ + 8H2O + 5e-Now, the number of oxygen atoms is balanced.Step 7: Adding half-reactions, multiplying by integers (if necessary) to equalize the electrons in each half-reaction, and canceling species common to both sides.5NO2− + 2MnO4− + 16H+ → 5NO3− + 2Mn2+ + 8H2OFinally, the balanced equation of the redox reaction is:5NO2−+ 6H+ + 2MnO4− → 5NO3− + 2Mn2+ + 3H2O.
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True or false? Lymph vessels pick up a little bit of fluid that leaves your blood vessels every time you heart pumps.
Answer:
Explanation:
true.
Answer:
True. why?
Because the fluid that is lost through the blood vessels is restored but with no central pump or heart, this device releases and collects fluid.
help? for me? yes? please?
Answer:
2 Mg + O2 -> 2 MgO
is a synthesis reaction.
Which of the following is a FALSE statement? *
A. Heat moves through solids by by conduction.
B. Molecules move faster in warmer substances.
C. Warm water is denser than cold water.
D. Heat moves through liquids and gases by convection.
Answer:
C: Warm water is denser than cold water
Mass = 22g Volume = 2cm3 What is the Density?
Answer:
11g/cm^3
Explanation:
The density of a substance can be found By
density= Mass/volume
From the question
mass = 22 g
volume = 2 cm³
22/2=11
We have the final answer as
11 g/cm³
Answer:
11g/cm^3
Explanation:
It is because the constant formula density is
Density=Mass/ Volume.
Therefore we have
Density=22g/2cm^3
Which is = 11g/cm^3
What is the mass of sulfuric acid, h2so4, that is contained in 2.0L of a 5.85 M solution ?
A- 1147.44 grams
B- 0.12 grams
C- 11.7 grams
D- 98.07 grams
Answer:
A. 1147.44 grams
Explanation:
Multiply volume by molarity to get the moles of solution.
5.85 M = mol/2.0 L
mol = (5.85 M)(2.0 L)
mol = 11.7
There is 11.7 moles of H2SO4.
Convert to grams with molar mass.
11.7 mol H2SO4 x (98.076 g/1 mol) = 1147.49 g
Closest answer is A, 1147.44 g.
What is the process of old crust sinking back into a mantle?
Use the following words to fill in the text (can be used more than once).
Anode,Anode to cathode,Battery,Cathode,Chemical to electrical,Electrical to chemical,Electrolysis,Electroplating,Endothermic,Exothermic,N/A
Negative,No,Non-Spontaneous,Oxidation,Positive,Reduction,Spontaneous,Yes.
Answer:
VOLTIC CELL
Battery
Electrical to chemical
Endothermic
Electrolysis
Chemical to electrical
Anode
Anode to cathode
Battery
Endothermic
Negative
Enthomeric
Reduction
Oxidation
ELECTRYOTIC CELL
No
Non spontaneous
positive
negative
yes
Anode
Electrical to chemical
Anode to cathode
Exothermic
positive
oxidation
Electroplating
Chemical to electrical
For practice, the molecular formula for ascorbic acid is C6H8O6, and you used 6 g in this experiment, calculate the molarity of the ascorbic acid. Now calculate the concentration in moles per drop (assume 1 mL = 20 drops).
I do not want the answer. I want to know how to get the answer please. Just explain to me how to figure this out.
Answer:
0.68 M
Explanation:
The concentration of a solute refers to Molarity= [tex]\frac{moles}{Liter}[/tex]
We need to find moles, we can do that by using stoichiometry to convert 6g > moles using molar mass.
Molar mass of C6H8O6= 176 g/mol; this is our key to converting the amount we have to moles.
6g C6H8O6 x [tex]\frac{1 mol}{176 g}[/tex]C6H8O6 = 0.034 mol C6H8O6
Now we need our volume per drop.
We know 1000 mL=1 L
We have 20 drops per 1 mL, so 1 mL / 20 = 0.05 mL per drop. But we need Liters, so we'll convert it to Liters.
0.05 mL x [tex]\frac{1 L}{1000mL}[/tex] = 0.00005L
Now we can solve for Molarity.
Molarity of C6H8O6 = [tex]\frac{0.034 mols}{0.00005 L}[/tex] = 0.068M of C6H8O6
Shaun's doctor has recommended that he consume 200 calories of protein a day. Shaun consumed 176 calories of protein in one day. How much more protein does he need to consume to reach the required amount? (1 gram of protein = 4 calories)
Answer: Shaun must consume 6 more g of protein to reach the required amount.
Explanation:
Given : Recommended amount of protein = 200 calories
Now 4 calories = 1 gram of protein
Thus 200 calories = [tex]\frac{1}{4}\times 200=50[/tex] g of protein.
Calories consumed per day by Shaun = 176 calories
176 calories = [tex]\frac{1}{4}\times 176=44[/tex] g of protein.
Thus Shaun need to consume = (50-44) g = 6 of protein to reach the required amount.
Shaun must consume 6 more g of protein to reach the required amount.
What would it mean if a forecaster claimed, “There will be east winds of 25 to 30 mph
Answer:
there will be winds moving from 25miles per hour to 30 miles per hour towards the east
which set of elements is arranged in order of increasing electronegativity? A. cl < f < se < as B. se < f < as < cl C. f < cl < se < as D. as < se < cl < f
f < cl < se < as (fluorine, chlorine, selenium, arsenic), which arranges the elements in increasing order of electronegativity.
Hence, the correct option is C.
Electronegativity is a measure of an atom's ability to attract electrons towards itself when it is part of a chemical bond. It generally increases as you move across a period from left to right and decreases as you move down a group in the periodic table.
A. cl < f < se < as (chlorine, fluorine, selenium, arsenic):
Chlorine (Cl) has a higher electronegativity than fluorine (F), selenium (Se), and arsenic (As). Therefore, it is arranged correctly that chlorine has the highest electronegativity, followed by fluorine, selenium, and then arsenic.
B. se < f < as < cl (selenium, fluorine, arsenic, chlorine):
This arrangement is not in increasing order of electronegativity. Fluorine (F) has the highest electronegativity among these elements, followed by chlorine (Cl), then arsenic (As), and selenium (Se) has the lowest electronegativity.
C. f < cl < se < as (fluorine, chlorine, selenium, arsenic):
This arrangement is correct. Fluorine (F) has the highest electronegativity, followed by chlorine (Cl), then selenium (Se), and arsenic (As) has the lowest electronegativity.
D. as < se < cl < f (arsenic, selenium, chlorine, fluorine):
This arrangement is not in increasing order of electronegativity. Fluorine (F) has the highest electronegativity among these elements, followed by chlorine (Cl), then selenium (Se), and arsenic (As) has the lowest electronegativity.
Therefore, the correct answer is option C: f < cl < se < as (fluorine, chlorine, selenium, arsenic), which arranges the elements in increasing order of electronegativity.
Hence, the correct option is C.
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.(Part A)
Write balanced net ionic equation for the following reaction:
Fe(OH)3(s)+H2SO4(aq)→?
Express your answer as a chemical equation. Identify all of the phases in your answer.
(Part B)
Write balanced net ionic equation for the following reaction:
HClO3(aq)+Na(OH)(aq)→?
Note that HClO3 is a strong acid.
Express your answer as a chemical equation. Identify all of the phases in your answer.
Part A: [tex]Fe(OH)_{3} (s)[/tex] + 3H+(aq) → [tex]Fe^3+(aq)[/tex]+ [tex]3H_{2}O(l)[/tex](solid [tex]Fe(OH)_{2}[/tex] reacts with aqueous H+ to form [tex]Fe^3+[/tex] ions and water). Part B: H+(aq) + [tex]ClO_{3}^-[/tex](aq) + Na+(aq) +[tex]OH^-(aq)[/tex]→ Na+(aq) + [tex]ClO_{3}^-(aq)[/tex] + [tex]H_{2}O(l)[/tex](H+ reacts with [tex]OH^-[/tex] to form water in the presence of Na+ and [tex]ClO_{3}^-[/tex] ions).
Part A:
The balanced net ionic equation for the reaction between solid [tex]Fe(OH)_{3}[/tex] and aqueous [tex]H_{2}SO_{4}[/tex]is:
[tex]Fe(OH)_{3} (s)[/tex] + 3H+(aq) + [tex]3SO_{4}^2-(aq)[/tex] → [tex]Fe^3+(aq)[/tex] + [tex]3SO_{4}^2-(aq)[/tex] +[tex]3H_{2}O(l)[/tex].
In this reaction, the [tex]Fe(OH)_{3}[/tex] solid reacts with the H+ ions from [tex]H_{2}SO_{4}[/tex]to form[tex]Fe^3+[/tex] ions and water molecules. The sulfate ions [tex](SO_{4}^2-)[/tex]are spectator ions and do not participate in the net reaction. The phases are indicated by (s) for solid, (aq) for aqueous, and (l) for liquid.
Part B:
The balanced net ionic equation for the reaction between [tex]HClO_{3}(aq)[/tex] and NaOH(aq) is:
H+(aq) + [tex]ClO_{3}^-(aq)[/tex]+ Na+(aq) + [tex]OH^-(aq)[/tex] → Na+(aq) + [tex]ClO_{3}^-(aq)[/tex] + [tex]H_{2}O(l)[/tex].
Here, the H+ ion from [tex]HClO_{3}[/tex] reacts with the [tex]OH^-[/tex] ion from NaOH to form water molecules. Sodium ion (Na+) and chlorate ion ([tex]ClO_{3}^-[/tex]) are spectator ions. The phases are indicated by (aq) for aqueous and (l) for liquid.
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To remove a tight-fitting lid from a jar, Megan runs the lid under hot water.
What happens to the Jar lid when its temperature increases?
A.The temperature increases, and the lid expands
B.The temperature decreases, and the lid contracts.
C.The potential energy increases, and the lid contracts.
D.The potential energy decreases, and the lid expands
Answer:
A
Explanation:
Which of the following explanations accounts for the fact that the ion-solvent interaction is greater for Li+ than for K+?
The ionization energy of Li is higher than that for K.
Li has a lower density than K.
Li+ is of a lower mass than K+
Li+has a smaller ionic radius than K+
Li reacts with water more slowly than K.
The most appropriate explanation among the given options for the fact that the ion-solvent interaction is greater for Li⁺ than for K⁺ is that "Li⁺ has a smaller ionic radius than K⁺." Option D is correct.
Ionic radius refers to the size of an ion, specifically the distance between the nucleus and the outermost electron shell. Smaller ions tend to have stronger interactions with solvent molecules because the charge is more concentrated in a smaller space, leading to a higher electrostatic attraction between the ion and the surrounding solvent molecules.
In this case, Li⁺ has a smaller ionic radius compared to K⁺. As a result, Li⁺ has a stronger interaction with the solvent molecules, which leads to a greater ion-solvent interaction.
Hence, D. is the correct option.
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--The given question is incomplete, the complete question is
"Which of the following explanations accounts for the fact that the ion-solvent interaction is greater for Li⁺ than for K⁺? A) The ionization energy of Li is higher than that for K. B) Li has a lower density than K. C) Li⁺ is of a lower mass than K⁺ D) Li⁺ has a smaller ionic radius than K⁺ E) Li reacts with water more slowly than K."--
Given the balanced chemical equation: 2 H2 + O2 → 2 H20. Choose
the description below that most accurately describes the reaction.
O 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water
O 2 moles of water react with 1 mole of hydrogen to form 2 moles of oxygen
O equal moles of hydrogen and oxygen react to form an equal amount of
water
O 4 moles of hydrogen react with 2 moles of oxygen to form 4 moles of water
Answer:
The answer is
2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water
it takes 45 hours for a 6.00 mg sample of sodium-24 to decay to 0.750 mg. what is the half-life of sodium-24?
To find the half-life of sodium-24, we can use the formula for exponential decay:
N(t) = N₀ * (1/2)^(t / t₁/₂)
Where:
N(t) is the amount of the substance remaining at time t
N₀ is the initial amount of the substance
t is the time elapsed
t₁/₂ is the half-life of the substance
Given:
N₀ = 6.00 mg (initial amount)
N(t) = 0.750 mg (amount after 45 hours)
Plugging in these values, we can solve for t₁/₂:
0.750 mg = 6.00 mg * (1/2)^(45 / t₁/₂)
Dividing both sides of the equation by 6.00 mg:
(0.750 mg) / (6.00 mg) = (1/2)^(45 / t₁/₂)
0.125 = (1/2)^(45 / t₁/₂)
To eliminate the base of 1/2, we can take the logarithm of both sides:
log₂(0.125) = 45 / t₁/₂ * log₂(1/2)
Using the property log₂(a^b) = b * log₂(a):
log₂(0.125) = 45 * log₂(1/2) / t₁/₂
We know that log₂(1/2) = -1, so we can simplify the equation further:
log₂(0.125) = -45 / t₁/₂
Now, we can solve for t₁/₂:
t₁/₂ = -45 / log₂(0.125)
Using a calculator:
t₁/₂ ≈ -45 / (-3) ≈ 15
The half-life of sodium-24 is approximately 15 hours.
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If the concentration of OH in a saturated solution of Ca(OH)₂ is 5.2 x 10-² M, is the solubility of Ca(OH)2 greater than 0.1g/100.mL? (Molar mass of Ca(OH)2 is 74.1 g/mol) a. Same b. No C. Yes
Comparing the calculated solubility of Ca(OH)₂ (1.924 g/100 mL) to the given value of 0.1 g/100 mL, we can conclude that the solubility of Ca(OH)₂ is greater than 0.1 g/100 mL.
Option (c) is correct.
To determine if the solubility of Ca(OH)₂ is greater than 0.1 g/100 mL based on the given concentration of OH⁻, we need to calculate the solubility of Ca(OH)₂ using the provided concentration.
The balanced chemical equation for the dissociation of Ca(OH)₂ is:
Ca(OH)₂ ⇌ Ca²⁺ + 2OH⁻
From the equation, we can see that for every mole of Ca(OH)₂ that dissolves, two moles of OH⁻ ions are produced. Therefore, the concentration of OH⁻ is twice the concentration of Ca(OH)₂.
Given that the concentration of OH⁻ is 5.2 x 10⁻² M, the concentration of Ca(OH)₂ can be calculated by dividing the concentration of OH⁻ by 2:
Ca(OH)₂ concentration = (5.2 x 10⁻² M) / 2 = 2.6 x 10⁻² M
To determine the solubility of Ca(OH)₂ in grams per 100 mL, we can use the molar mass of Ca(OH)₂:
Solubility of Ca(OH)₂ = (2.6 x 10⁻² M) x (74.1 g/mol) = 1.924 g/100 mL
Comparing the calculated solubility of Ca(OH)₂ (1.924 g/100 mL) to the given value of 0.1 g/100 mL, we can conclude that the solubility of Ca(OH)₂ is greater than 0.1 g/100 mL.
Therefore, the correct answer is c) Yes.
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Newton's third law of motion is summarized as the law of action-reaction. For every action, there is an equal but opposite reaction. Which example best illustrates the application of Newton's law of action-reaction? Choose all that apply.
Answer:
When we stand on the floor, we apply a force on the floor surface in the downward direction and in return the floor also exerts an upward and equal force on us.
Explanation:
Newton' third law is vey famous and it states that for each and every action, there applies an equal but opposite reaction. Thus the action force and the reaction force always acts on pairs. But they does not contribute to the motion of the object.
One such example that illustrates the action and reaction force from Newton's law is when we stand on the floor we exert a force on the floor surface in downward direction. The floor surface also exerts an opposite and equal force on us in the upward direction.
Calculate the number of moles in 7.04 g of lead (II) nitrate, Pb(NO3)2
Can Someone help me covert this?!
Answer:
moles Pb(NO₃)₂ ≅ 0.0213 mole (three sig. figs.)
Explanation:
converting...
grams to moles => divide by formula weight
moles to grams => multiply by formula weight
for this problem ...
moles Pb(NO₃)₂ = 7.04g / 331.2g/mol = 0.021256 (calculator answer)
≅ 0.0213 mole (three sig. figs.)
Mole measure the number of elementary entities of a given substance that are present in a given sample. Therefore, the number of moles in 7.04 g of lead (II) nitrate, Pb(NO₃)₂ is 0.0212moles.
What is mole?The SI unit of amount of substance in chemistry is mole. The mole is used to measure the quantity or amount of substance. We know one mole of any element contains 6.022×10²³ atoms which is also called Avogadro number.
Mathematically,
mole =given mass ÷ molar mass
Molar mass of 1 mole of Pb(NO₃)₂ = 331.2g/mol
given mass of Pb(NO₃)₂=7.04g
Substituting the values in above formula we get
moles of Pb(NO₃)₂ = 7.04g / 331.2g/mol
moles of Pb(NO₃)₂ = 0.0212moles
Therefore, the number of moles in 7.04 g of lead (II) nitrate, Pb(NO₃)₂ is 0.0212moles.
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if i have 340 ml of a 0.5 m nabr solution, what will the concentration be if i add 500 ml more water to it?
The final concentration of the solution is determined as 0.59 M.
What is the final concentration of the solution?The concentration of a solution or molarity of a solution is defined as the ratio of number of moles of solute to the liters of the solution.
Molarity, M = n / V
where;
n is the number of moles of the soluteV is the volume of the solutionThe final concentration of the solution is calculated as follows;
the number of moles = 0.5 m
the total volume after adding 500 ml of water = 340 ml + 500 ml = 840 ml = 0.84 L
Molarity = 0.5 m / 0.84 L
Molarity = 0.59 M
Thus, the final concentration of the solution is determined as 0.59 M.
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What is a net ionic equation? A. An equation that shows only those particles involved in the reaction, and that is balanced for mass but not for charge B. An equation that shows only those particles involved in the reaction, and that is balanced for both mass and charge C. An equation that describes the flow of oxygen in a combustion reaction D. An equation that shows the cations as reactants and the anions as products
Answer:
B. An equation that shows only those particles involved in the reaction, and that is balanced for both mass and charge
Explanation:
A net ionic equation is a chemical equation that will only show the elements, compounds, and ions that are directly involved in the chemical reaction. It comes from the balanced molecular equation and should be balanced by both mass and charge.
cbr4 has a higher vapor pressure at the same temperature than ccl4. T/F?
CBr₄ has a higher vapor pressure at the same temperature than CCl₄. The statement is False.
The vapor pressure of a liquid is the pressure exerted by the vapor of the liquid in equilibrium with the liquid. The vapor pressure of a liquid depends on the strength of the intermolecular forces in the liquid. The stronger the intermolecular forces, the lower the vapor pressure.
In the case of CCl₄ and CBr₄, the intermolecular forces are London dispersion forces. The size of the atom or molecule affects the strength of the London dispersion forces. The larger the atom or molecule, the stronger the London dispersion forces.
Bromine is larger than chlorine, so the intermolecular forces in CBr₄ are stronger than the intermolecular forces in CCl₄. Therefore, CCl₄ will have a higher vapor pressure than CBr₄ at the same temperature.
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Our body is also a good conductor
of electricity because of the
presence of ______.
Please answer it this is my last question on my test i'll give you 20 points to answer?
Question 4 (2 points)
Manganese-56 decays by beta emission and has a half life of 2.6 hours. If a sample
is analyzed and found to contain 6.25% parent isotopes and 93.75% daughter
isotopes, what must the age of the sample be?
Answer:
10.4 hours
Explanation:
Now from;
0.693/t1/2 = 2.303/t log (No/N)
t1/2= half life of the radioactive isotope
t = age of the radioactive isotope
No= Original amount of radioactive isotope
N= amount of radioactive isotope left after time t
Substituting values;
N = 0.0625No
0.693/2.6 = 2.303/t log (No/0.0625No)
0.693/2.6 = 2.303/t log (1/0.0625)
0.267 = 2.303/t * log 16
0.267 = 2.773/t
t = 2.773/0.267
t = 10.4 hours
4. Identify What do scientists call a substance that forms during a chemical reaction?
5. Identify On what side of the equation are reactants found?
Answer:
4. Reactant
5. Left
I hope this helps ❤️❤️
In a food chain, energy does NOT flow directly from - F producer to decomposer G producer to consumer H consumer to decomposer J consumer to producer
Answer:
producer to decomposer
Explanation:
This is because in a food chain , energy flow from one trophic level to another. The producer which include plants are the source of energy which they manufacture good in the presence of light energy from sun. Energy flow directly from the producer to the primary consumer which are heterotrophs that feed on plants. Energy flow from consumer to decomposer after the consumer died and it is decayed.