parts per million (ppm), parts per billion (ppb) and parts per trillion (ppt) are the most commonly used terms to describe very small amounts of substances. a ppm of a chemical in water means that, in a million units of water, there would only be one unit of the chemical.
The smallest form of matter that still retains the properties of an element
Answer:
atom
Explanation:
the atom is the smallest form.
Label the parts of the electric circuit that best correspond to the heart, arteries, veins, and cells.
Answer:
1 ➡️ Cells
2 ➡️ Arteries
3 ➡️ Veins
4 ➡️ Heart
Explanation:
The parts of the electric circuit that best correspond to the heart, arteries, veins, and cells have been properly labeled.
The circulatory system involves the transportation of nutrients, oxygen and water by blood to other the parts of the body.
From the electric circuit, we see that arteries transport blood away from the heart to the other cells in the body. The veins actually return the blood back to the heart from the cells. The heart pumps the blood
The electric circuity diagram has the label 1 bulb analogous to cell, label 2 analogous to arteries, label 3 analogous to veins, and label 4 cell analogous to heart.
What is an electric circuit?The electric circuit has been given as the power source and the conducting wires that allows the flow of the current in the circuit.
In the human body, the heart has been transported the oxygenated blood through the arteries to the cell and carried the deoxygenated blood from the cells back to the heart via veins.
In the circuit, the battery has been the source of the power/blood. The current has been carried from the heart to the cell/bulb through the arteries labeled, 2, and transported back to the battery via veins labeled 3.
Learn more about the electric circuits, here:
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5. What gases was produced from decomposing hydrogen peroxide? What non-gaseous product formed from the reaction
Answer:
H2O
Explanation:
The equation for the decomposition of hydrogen peroxide is shown below;
2H2O2(l)-----> 2H2O(l) + O2(g)
Hence, the decomposition of H2O2 yields oxygen gas and water. Water is a non gaseous product of the reaction as clearly seen in the equation above.
3. A student took a calibrated 200.0 gram mass, weighed it on a laboratory balance, and
found it read 196.5 9. What was the student's percent error?
Answer:
The answer is 1.71 %Explanation:
The percentage error of a certain measurement can be found by using the formula
[tex]P(\%) = \frac{error}{actual \: \: number} \times 100\% \\ [/tex]
From the question
actual mass = 200 g
error = 200 - 196.59 = 3.41
We have
[tex]p(\%) = \frac{3.41}{200} \times 100 \\ = 1.705[/tex]
We have the final answer as
1.71 %Hope this helps you
If 5.00g of iron metal is reacted with 0.950g of Cl2 gas, how many grams of ferric chloride (FeCl3) will form?
Answer:
1.45g of FeCl3
Explanation:
The equation of the reaction is given as;
2Fe + 3Cl2 --> 2FeCl3
2 mol of Fe reracts with 3 mol of Cl2 to form 2 mol of FeCl3
Upon converting to mass using;
Mass = Number of moles * Molar mass
( 2 * 55.85 = 111.7g ) of Fe reacts with ( 3 * 71 = 213g ) of Cl2 to form ( 2 * 162.2 = 324.4g) of FeCl3
Cl2 is the limiting reactant as it determines how much of FeCl3 is formed
213g of Cl2 = 324.4g of FeCl3
0.950g of Cl2 = x
x = (0.950 * 324.4 ) / 213
x = 1.45g of FeCl3
To what volume would you need to dilute 200 mL of a 5.85M solution of Ca(OH)2 to make it a 1.95M solution?
Answer: 600 mL
Explanation:
Given that;
M₁ = 5.85 m
M₂ = 1.95 m
V₁ = 200 mL
V₂ = ?
Now from the dilution law;
M₁V₁ = M₂V₂
so we substitute
5.85 × 200 = 1.95 × V₂
1170 = 1.95V₂
V₂ = 1170 / 1.95
V₂ = 600 mL
Therefore final volume is 600 mL
Which of the following substances will form a base when dissolved in water?
Question 14 options:
A)
HCl
B)
HBr
C)
KOH
D)
SO2
Answer:
Its D) SO2
Explanation
Did the test
Which is a chemical property of milk
A. Milk has a ph ranging from 6.4 to 6.8
B. Milk spoils when left unrefrigerated
C. Milk boils at about 212F
D. Milk curdles when mixed with vinegar
Answer:
C. Milk boils at about 212F
Explanation:
The principal constituents of milk are water, fat, proteins, lactose (milk sugar) and minerals (salts). Milk also contains trace amounts of other substances such as pigments, enzymes, vitamins, phospholipids (substances with fatlike properties), and gases.
How many grams of NaCl are needed to prepare 50.0 grams of a 35.0% salt solution?
17.5 grams of NaCl are needed to prepare 50.0 grams of a 35.0% salt solution.
What is Solution?
A solution is a homogenous mixture composed of two or more substances. In a solution, the components are evenly distributed throughout the mixture, resulting in a uniform appearance and properties.
The substance that is present in the largest amount is called the solvent, while the other substances present in lesser amounts are called solutes. When the solute dissolves in the solvent, the resulting mixture is called a solution.
To calculate the grams of NaCl needed to prepare a 35.0% salt solution, we can use the formula:
grams of NaCl = (percent salt / 100) x grams of solution
grams of NaCl = (35.0 / 100) x 50.0
grams of NaCl = 0.35 x 50.0
grams of NaCl = 17.5
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A sailor on a trans-Pacific solo voyage notices one day that if he puts 375. mL of fresh water into a plastic cup fresh water weighing 15.0 g, the cup floats in the seawater around his boat with the fresh water inside the cup at exactly the same level as the seawater outside the cup.
Required:
Calculate the amount of salt dissolved in each liter of seawater.
Answer:
Amount of salt dissolved in each liter of seawater = 40 g
Explanation:
According to Archimedes principle, a body will float in a fluid if the upthrust experienced by a body is equal to the to the weight of the body.
Also, the volume of seawater displaced equals the volume of freshwater in the cup.
From the above principle, since the freshwater and cup floats in the seawater, their combined weight equals the upthrust.
Therefore, mass of equal volume of displaced seawater = mass of freshwater + mass of cup
Mass of freshwater = density of freshwater * volume
density of freshwater = 1 g/mL; volume = 375 mL
mass of freshwater = 375 mL * 1 g/mL = 375 g
mass of seawater = 375 + 15 = 390 g
mass of salt in 375 mL seawater = mass of seawater - mass of freshwater
mass of salt = (390 - 375) g = 15 g
Since 15 g of salt are dissolved in 375 mL seawater, mass of salt in 1 L of seawater =(1000 mL/ 375) * 15g = 40 g
Therefore, amount of salt dissolved in each liter of seawater = 40 g
which element Shows very similar chemical properties to barium?
What does chemical equations and chemical formulas have in common?
Answer:
Chemical symbols refer to chemical elements only. They do not necessarily refer to atoms of that element, but also to ions.
Explanation:
4. CHALLENGE Suppose you had a mixture of sand and small,
hollow beads. How might you separate the mixture?
I'm not sure if this is the answer but maybe oil.
1) The speed constant for the second order reaction in the gas phase
It varies with the temperature according to the table below. Calculate the activation energy for the process, according to Arhhenius' equation
Answer:
41.7 kJ/mol
Explanation:
ln(k) = ln(A) − Eₐ/(RT)
Pick any two points. I'll choose 100°C and 400°C.
When T = 100°C = 373 K, k = 1.10×10⁻⁹ L/mol s:
ln(1.10×10⁻⁹) = ln(A) − Eₐ/(R × 373)
When T = 400°C = 673 K, k = 4.40×10⁻⁷ L/mol s:
ln(4.40×10⁻⁷) = ln(A) − Eₐ/(R × 673)
Subtract the two equations and solve:
ln(4.40×10⁻⁷) − ln(1.10×10⁻⁹) = -Eₐ/(R × 673) + Eₐ/(R × 373)
5.991 = 0.00120 Eₐ/R
Eₐ/R = 5013.4
Eₐ = 41700 J/mol
Eₐ = 41.7 kJ/mol
Select the term that matches each definition:
a) A decrease in the solubility of an ionic compound as a result of the addition of a common ion.
b) The mass of a salt in grams that will dissolve in 100 mL of water.
c) A solution that has dissolved the maximum amount of a compound at a given temperature. Any further addition of salt will remain undissolved.
d) The product of the molarities of the dissolved ions, raised to a power equal to the ion's coefficient in the balanced chemical equation.
e) The maximum number of moles of a salt that will dissolve in 1 L of solution.
*** Answer options for all questions: ***
- Solubility
- Molar Solubility
- Solubility product constant
- Common ion effect
- Saturated Solution
Answer:
a) common ion effect
b) solubility
c) saturated solution
d) solubility product constant
e) molar solubility
Explanation:
When a substance, say BA2 is dissolved in a solution and another substance CA2 is dissolved in the same solution. The solubility of BA2 is decreased due to the addition of CA2. This is known as common ion effect.
The mass of a substance that will dissolve in a given Volume of solvent is known as it's solubility.
The molar solubility is the amount of moles of solvent that dissolves in 1 dm^3 of solvent.
A solution that contains just as much solute as it can normally hold at a given temperature is known as a saturated solution.
Lastly, the product of molar solubilites raised to the power of the molar coefficient is know as the solubility product constant.
The correct matches and their explanation are:
a) A decrease in the solubility of an ionic compound as a result of the addition of a common ion: Option 4. common ion effect
It relates to the equilibrium effect that occurs due to the addition of common ions.b) The mass of salt in grams that will dissolve in 100 mL of water: Option 1. solubility
Solubility is the property of solute to form a solution with the solvent of another substance.c) A solution that has dissolved the maximum amount of a compound at a given temperature. Any further addition of salt will remain undissolved: Option 5. saturated solution
When the solution cannot accommodate any more addition of solute of a substance is called a saturated solution.d) The product of the molarities of the dissolved ions, raised to a power equal to the ion's coefficient in the balanced chemical equation: Option 3. solubility product constant
It is an equilibrium constant for the solid solute dissolved in the solution.e) The maximum number of moles of a salt that will dissolve in 1 L of solution: Option 2. molar solubility
Before the saturation of a solution, the amount of solute it can dissolve is called molar solubility.To learn more about molar solubility and common ion effect follow the link:
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how many moles are in a 4.2 gram gold sample
Which of the following represents a species with 16 protons and 18 electrons? A) Ar B) S C) S²⁻ D) Si⁴⁻ E) S²⁺ Question 5 of 40
Answer:S²⁻
Explanation:
Sulphur is in group 16. The atomic number of sulphur is 16.
However, sulphur can accept two electrons to form the sulphide ion S²⁻ which contains 18 electrons, hence the answer provided above.
g Calculate the mass percent of sodium bicarbonate in the solution that has 7.00 g of solution and 0.052 Kg of water.
Answer:
11.86%
Explanation:
First, we convert both solvent and solute to the same unit, say, kg. We have.
Mass of Sodium Bicarbonate = 7g = 7/1000 kg = 0.007 kg
Mass of water = 0.052 kg.
Formula for the mass percent is
% of sodium bicarbonate = [(mass of sodium bicarbonate) / (mass of total solution) * 100%]
Total mass of solution = 0.007 + 0.052 Total mass of solution = 0.059
% of sodium bicarbonate = 0.007 / 0.059 * 100%
% of sodium bicarbonate = 11.86%
Therefore, the mass percent of sodium bicarbonate I'm the solution is 11.86%
I don’t really understand, if anybody can help I’ll really appreciate it ! Thank you.
Answer:
175
Explanation:
How to calculate calories
Answer:If you are sedentary (little or no exercise) : Calorie-Calculation = BMR x 1.2.
If you are lightly active (light exercise/sports 1-3 days/week) : Calorie-Calculation = BMR x 1.375.
Explanation:
how many moles of h2 can be made from the complete reaction of 3.5 moles of al?
Given: 2Al+6HCL 2Alcl3+3h2
Answer:
From the given equation, we can see that for every 2 moles of Al, we get 3 moles of H2
So, we can say the the number of moles of H2 is 3/2 times the number of moles of Al
We are given the number of moles of Al and we have to find the number of moles of H2
We have deduced the relationship:
Moles of Al * 3 / 2 = Moles of H2
Replacing the variables with given values
3.5 * 3 / 2 = Moles of H2
Moles of H2 = 5.25 moles
8 You are given 20.00g of a dry mixture of sand and table salt.
After adding water and filtering, you are left with wet sand on
the filter paper. The filter paper and sand is then dried and the
mass of the dry sand alone is found to be 5.00g. What was the
% sand in the original mixture?
Answer:
[tex]\%sand=25\%[/tex]
Explanation:
Hello.
In this case, given the mass of the mixture, we can define it in terms of the mass of sand and table salt as shown below:
[tex]m_{sand}+m_{salt}=20.00g[/tex]
Moreover, as after filtering, the mass of dry sand turns out 5.00 g, we can compute the % sand in the original mixture by dividing this value over the mass of the mixture as shown below:
[tex]\%sand=\frac{m_{sand}}{m_{mixture}}*100\%\\ \\\%sand=\frac{5.00g}{20.00g}* 100\%\\\\\%sand=25\%[/tex]
Best regards!
How to separate given mixture?
Answer:
Chromatography involves solvent separation on a solid medium.
Distillation takes advantage of differences in boiling points.
Evaporation removes a liquid from a solution to leave a solid material.
Filtration separates solids of different sizes.
Explanation:
The solubility of silver(I)phosphate at a given temperature is 2.43 g/L. Calculate the Ksp at this temperature. After you calculate the Kspvalue, take the negative log and enter the (pKsp) value with 2 decimal places.
Answer:
Kps = 3.07 x 10⁻⁸
pKsp= 7.51
Explanation:
First, we calculate the molar solubility of silver(I)phosphate (Ag₃PO₄) from the solubility in g/L by using its molar mass (418.6 g/mol):
2.43 g/L x 1 mol/418.6 g = 5.8 x 10⁻³ mol/L= s
Now, we have to write the ICE chart for the aqueous equilibrium of Ag₃PO₄ as follows:
Ag₃PO₄(g) ⇄ 3 Ag⁺(aq) + PO₄³⁻
I 0 0
C +3s +s
E 3s s
Ksp = [Ag⁺]³[PO₄³⁻]= (3s)³s= 27s⁴
Since s=5.8 x 10⁻³ mol/L, we calculate Ksp:
Ksp= 27(5.8 x 10⁻³ mol/L)⁴= 3.07 x 10⁻⁸
The pKsp value is:
pKsp= - log Ksp = -log (3.07 x 10⁻⁸) = 7.51
twelve grams of sodium chloride wwere dissolved in 52 ml (52g) of distilled water, calculate the % sodium chloride in the solution
Answer:
The mass of sodium chloride in the mixture is 18.75%
Explanation:
Here, we want to calculate the percentage of sodium chloride in the mixture.
The total mass of the mixture is 52 g + 12 g = 64 g
So the percentage mass of sodium chloride will be;
mass of sodium chloride/ Total mass * 100%
That will be: 12/64 * 100 = 18.75%
Which is the product of that reaction
Answer:
B
Explanation:
a change of matter is a physical change
True or False
Answer:
true
Explanation:
hope it helps
Answer:
I'm pretty sure it's true
Explanation:
20 characters
A student was performing a separation of a mixture of organic compounds. The final step of the process involved a filtration of the analyte from an aqueous solution. After drying the filtered solid for a very short period time, they took the melting point of the compound. The measured melting point range of the compound was 106 – 113.8 0C, while the literature melting point of the compound was 122.3 0C. The above scenario is a very common one in organic labs.
1. Do you think their sample was pure?
2. If not, then what do you think could be the source of error.
3. How do you think this error can be minimized?
Answer:
1) No
2) The solvent contaminated the analyte
3) The solvent should be evaporated properly before washing and drying the analyte
Explanation:
During separation of organic compounds, solvents are used. These solvents are able to contaminate the analyte and lead to a large difference in melting point of solids obtained.
However, the error can be minimized by evaporating the solvent before washing, drying and melting point determination of the solid.
When nitrogen, oxygen, fluorine, sodium, magnesium and aluminum ionize, they all will have:
a. different electron configuration from each other.
b. an unchanged electron configuration.
c. the same charge.
d. the same electron configuration (isoelectronic) as neon.
[Definition: The word isoelectronic means that when you write out the electron configuration they are the same. An exam would be He and Li whereby both of them have 2 electrons and therefore they are both are 1s2 in their electron configurations.]
Answer: d. the same electron configuration (isoelectronic) as neon.
Explanation:
Isoelectronic species are defined as the molecules which have the same number of electrons.
Atomic number of nitrogen is 7 and thus has 7 electrons. Nitrogen has electronic configuration of 2,5 and thus can gain 3 electrons and thus [tex]N^{3-}[/tex] will have electronic configuration of 2,8 ( same as that of neon)
Atomic number of oxygen is 8 and thus has 8 electrons. Oxygen has electronic configuration of 2,6 and thus can gain 2 electrons and thus [tex]O^{2-}[/tex] will have electronic configuration of 2,8 ( same as that of neon)
Atomic number of flourine is 9 and thus has 9 electrons. Flourine has electronic configuration of 2,7 and thus can gain 1 electron and thus [tex]F^{-}[/tex] will have electronic configuration of 2,8 ( same as that of neon)
Atomic number of sodium is 11 and thus has 11 electrons. Sodium has electronic configuration of 2,8,1 and thus can lose 1 electron and thus [tex]Na^{+}[/tex] will have electronic configuration of 2,8 ( same as that of neon)
Atomic number of magnesium is 12 and thus has 12 electrons. Magnesium has electronic configuration of 2,8,2 and thus can lose 2 electrons and thus [tex]Mg^{2+}[/tex] will have electronic configuration of 2,8 ( same as that of neon)
Atomic number of aluminium is 13 and thus has 13 electrons. Aluminium has electronic configuration of 2,8,3 and thus can lose 3 electrons and thus [tex]Al^{3+}[/tex] will have electronic configuration of 2,8 ( same as that of neon)
the egents erosion and explain how each of them causes erosion
Answer:
Wind, water, gravity, and ice
Explanation:
Water can erode soil material. Especially if the soil is bare, dry and erodible erosion via rain particles can occur. Wind is another factor that cause soil erosion. Dry soil particles( Especially if they are fine) can move to other areas if wind exists. Ice is another issue on erosion. Ice, when it is melting, can carry soil particles.