Answer:
A sample of an ideal gas has a volume of 2.21 L at 279 K and 1.01 atm. Calculate the pressure when the volume is 1.23 L and the temperature is 299 K.
You need to apply the ideal gas law PV=nRT
You have the pressure, P=1.01 atm
you have the volume, V = 2.21 L
The ideal gas constant R= 0.08205 L. atm/ mole.K at 273 K
find n = PV/RT = (1.01 atm x 2.21 L / 0.08205 L.atm/ mole.K x 273 K)
n= 0.1 mole, Now find the pressure for n=0.1 mole, T= 299K and
L=1.23 L
P=nRT/V= 0.1mole x 0.08205 (L.atm/ mole.K x 299 k)/ 1.23 L
= 1.994 atm
Explanation:
Solid calcium chlorate decomposes to form solid calcium chloride and oxygen gas.
Write the balanced chemical equation for the reaction described. Phases are optional.
equation:
Answer:
Ca(ClO₃)₂(s) → CaCl₂(s) + 3O₂(g)
Explanation:
Chemical equation:
Ca(ClO₃)₂(s) → CaCl₂(s) + O₂(g)
Balance chemical equation:
Ca(ClO₃)₂(s) → CaCl₂(s) + 3O₂(g)
Step 1:
Ca(ClO₃)₂(s) → CaCl₂(s) + O₂(g)
Left hand side Right hand side
Ca = 1 Ca = 1
Cl = 2 Cl = 2
O = 6 O = 2
Step 2:
Ca(ClO₃)₂(s) → CaCl₂(s) + 3O₂(g)
Left hand side Right hand side
Ca = 1 Ca = 1
Cl = 2 Cl = 2
O = 6 O = 6
What is the process of cell eating called
Answer:
Phagocytosis
Explanation:
When a helium balloon rises in the air, it expands. If the volume of the balloon doubles, what happens to the density of the helium inside it?
a.The density decreases by half
b.The density doubles
c.The density triples
d.The density stays the same
convert 575.1 mmHg to atm
Answer:
= .7567105263
Explanation:
1 atm = 760 mmHg
575.1 mmHg (1 atm/760mmHg) = .7567105263 atm
How does temperature affect the copper (II) chloride equilibrium? Is the forward reaction (color changing from blue to green) endothermic or exothermic? Justify your choice with experimental evidence i.e color changes in the video for Part B.
Answer:
See explanation
Explanation:
A popular experiment that describes the effect of heat on the position of equilibrum is the change of colour when copper II chloride is heated.
As the solution is heated, it's colour changes from blue to green, this implies the the colour change (blue to green) is an endothermic process (equilibrum position shifts to the right with increase in temperature)
The equilibrum is represented by the equation;
[Cu(H2O)6]^2+(aq) + 4Cl^-(aq)<------>[CuCl4]^2-(aq) + 6H2O(l) ∆H=positive
The equilibrium mixture undergoing cooling or heating have colour changes. The temperature affects the colour of the products formed and the forward reaction is endothermic.
What are the equilibrium and the forward reactions?In the reaction copper (II) chloride or [tex]\rm CuCl_{4}[/tex] is the main species. The heat or the temperature affects the colour formation of copper (II) chloride as the equilibrium change affects the colouration of the product.
The heating of the solution affects the colour change from blue to the green of the reactant to products and the forward reaction shifts the equilibrium towards the right when the temperature is increased and is an endothermic reaction.
The reaction at the equilibrium can be shown as,
[tex]\rm [Cu(H_{2}O)_{6}]^{2+} (aq) + 4Cl^{-} (aq) \Leftrightarrow [CuCl_{4}]^{2-}(aq) + 6H_{2}O(l), \Delta H=positive[/tex]
Therefore, temperature changes the colouration and the forward reaction is an endothermic reaction.
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An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at a constant pressure of 35.0 atm and releases 74.6 kJ of heat. Before the reaction, the volume of the system was 8.20 L . After the reaction, the volume of the system was 2.80 L . Calculate the total internal energy change, ΔE, in kilojoules.
Answer:
ΔU = −55.45 kJ
Explanation:
From first law of thermodynamics in chemistry, we have;
ΔU = Q + W
where;
ΔU is change in internal energy
Q is the net heat transfer
W is the net work done
We are given;
Q = 74.6 kJ
But Q will be negative since heat is released
Thus;
ΔU = -74.6 kJ + W
We are given;
Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²
Volume before reaction; Vi = 8.2 L = 0.0082 m³
Volume after reaction; V_f = 2.8 L = 0.0028 m³
Now,
W = -P(V_f - V_i)
W = - 3546375(0.0028 - 0.0082)
W = 19.15 KJ
Thus;
ΔU = Q + W
ΔU = -74.6 kJ + 19.15 KJ =
ΔU = −55.45 kJ
A scientist discovers remnants of an organism on a slide under his microscope. He can only identify a few components: a large vacuole, ridged cell wall, and a chloroplast. Was the organism prokaryotic or eukaryotic? How do you know?
Answer:
Eukaryotic (a plant cell)
Explanation:
The presence of a chloroplast indicates that the cell has membrane-bound organelles. This is not a feature of prokaryotic cells - only eukaryotic cells. Therefore, the cell is eukaryotic. The presence of a large vacuole, chloroplast, and rigid cell wall suggests its a plant cell as plant cells are the only eukaryotes with these features.
The pKb values for the dibasic base B are pKb1=2.10 and pKb2=7.54. Calculate the pH at each of the points in the titration of 50.0 mL of a 0.60 M B(aq) solution with 0.60 M HCl(aq).
Complete Question
The pKb values for the dibasic base B are pKb1=2.10 and pKb2=7.54. Calculate the pH at each of the points in the titration of 50.0 mL of a 0.60 M B(aq) solution with 0.60 M HCl(aq).
(a) before addition of any HCl (b) after addition of 25.0 mL of HCl
Answer:
a The value is [tex]pH =12.81[/tex]
b [tex]pH = 11.9[/tex]
Explanation:
From the question we are told that
The first pKb value for B is [tex]pK_b_1 = 2.10[/tex]
The second pKb value for B is [tex]pK_b_2 = 7.54[/tex]
The volume is [tex]V = 50.0 mL =[/tex]
The concentration of B is [tex][B] = 0.60 M[/tex]
The concentration of [tex]C_A = 0.60 M[/tex]
Generally the reaction equation showing the first dissociation of B is
[tex]\ce{B_{(aq) } + H_2O _{(l)} <=> BH^+ _{(aq)} + OH^- _{(aq)} }[/tex]
Here the ionic constant for B is mathematically represented as
[tex]K_i = \frac{[BH^+] [OH^-]}{[B]}[/tex]
Let denot the concentration of [BH^+] as z and since [tex][BH^+] = [OH^-][/tex] then [tex][OH^-][/tex] is also z
So [B] = 0.60 - z
Here [tex]K_i[/tex] is ionic constant for the first reaction of a dibasic base B and the value is
[tex]K_i = 7.94 *10^{-3}[/tex]
So
[tex] 7.94 *10^{-3}= \frac{z^2}{ 0.60 - z}[/tex]
=> [tex]z^ 2 + 0.00794 z - 0.00476[/tex]
using quadratic formula to solve this equation
[tex]z = 0.0651[/tex]
Hence the concentration of [tex]OH^{-}[/tex] is [tex][OH^-] =0.0651[/tex]
Generally [tex]pOH = -log [OH^-][/tex]
=> [tex]pOH = -log (0.065)[/tex]
=> [tex]pOH = 1.187 [/tex]
Generally the pH is mathematically represented as
[tex]pH = 14 - 1.187[/tex]
[tex]pH =12.81[/tex]
Generally the volume of [tex]HCl[/tex] at the second dissociation of the base B is [tex] 50 mL [/tex]
The volume of the [tex]HCl[/tex] half way to the first dissociation of the base is 25mL
Now the pOH at half way to the first dissociation of the base is
[tex]pOH = -log(K_i)[/tex]
=> [tex]pOH = -log(0.00794)[/tex]
=> [tex]pOH = 2.100[/tex]
Generally the pH after addition of 25.0 mL of HCl is
[tex]pH = 14 - 2.100[/tex]\
=> [tex]pH = 11.9[/tex]
The first dissociation's equation is as follows:
[tex]B(aq) + H_2O(l) \leftrightharpoons BH^{+} (aq) + OH^{-}(aq) \\\\[/tex]
Constant of base ionization
[tex]\to K_{bl}=\frac{[BH^{+}][OH^{-}]}{[B]}\\\\ \to 7.94\times 10^{-3} = \frac{x\times x}{(0.95- x)} \\\\\to 7.94\times 10^{-3} = \frac{x^2}{(0.95- x)} \\\\\to x^2=7.94\times 10^{-3} (0.95-x) \\\\\to x^2=7.543\times 10^{-3} - 7.94\times 10^{-3} x) \\\\\to x^2=7.543\times 10^{-3} - 7.94\times 10^{-3} x) \\\\\to x = 0.0830\ M\\\\[/tex]
So,
[tex]\to [OH^{-}] = 0.0830\ M\\\\[/tex]
The second dissociation of the base equation is
[tex]BH^{+}\ (aq) + H_20\ (l) \leftrightharpoons BH_2^{2+}\ (aq) + OH^{-}\ (aq) \\\\[/tex]
Constant of base ionization
[tex]\to K_{bl}=\frac{[BH^{+}][OH^{-}]}{[B]}\\\\ \to 3.2 \times 10^{-8} =\frac{y \times (0.0830+y)}{(0.0830- y)}\\\\[/tex]
[tex]\to y= \frac{ 3.2 \times 10^{-8} \times (0.0830- y)}{ (0.0830+y)} \\\\ \to y= \frac{ 3.2 \times 10^{-8} \times (0.0830- y)}{ (0.0830+y)} \\\\ \to y = 3.2\times 10^{-8}[/tex]
So,
[tex]\to [OH^{-}] = 0.0830\ M \\\\\to pOH = 1.08 \\\\\to pH = 14.00 - pOH = 12.92\\\\[/tex]
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How do the test variables (independent variables) and outcome variables (dependent variables) in an experiment compare? A. The test variables (independent variables) and outcome variables (dependent variables) are the same things. B. The test variable (independent variable) controls the outcome variable (dependent variable). C. The test variable (independent variable) and outcome variable (dependent variable) have no affect on each other. D. The outcome variable (dependent variable) controls the test variable (independent variable).
Answer:
I'm on the exact same queston
Answer:
The test variable (independent variable) controls the outcome variable (dependent variable)
Explanation:
its right on study island
Which of the following is NOT a strong electrolyte?
:
Answers:
Na2SO4
KI
CaCl2
LiOH
C6H1206
Answer:
C6H1206
Explanation:
C6H12O6 is a monomer of carbohydrates also known as glucose, so it is not an electrolyte at all.
What is the percent error for the experiment if the actual density is
2.49g/mL but the experimental value is 1.47 g/mL?
Answer:
The answer is 40.96%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 density = 2.49g/mL
error = 2.49 - 1.47 = 1.02
We have
[tex]p(\%) = \frac{1.02}{2.49} \times 100 \\ = 40.96385542...[/tex]
We have the final answer as
40.96 %Hope this helps you
The normal boiling point of benzene is 80.1°C. What is its enthalpy of vaporization if the vapor pressure at 26.1°C is 100 torr?
The heat of vaporization of benzene is required.
The heat of vaporization of benzene is 33009 J/kg.
[tex]T_0[/tex] = Normal boiling point = 80.1+273.15 K
[tex]T_B[/tex] = Boiling point at given pressure = 26.1+273.15 K
[tex]R[/tex] = Gas constant = 8.314 J/mol K
[tex]P[/tex] = Pressure at given [tex]T_B[/tex] = 100 torr
[tex]\Delta H[/tex] = Heat of vaporization
From the Clausius–Clapeyron equation
[tex]\dfrac{1}{T_B}=\dfrac{1}{T_0}-\dfrac{R\ln(\dfrac{P}{P_0})}{\Delta H}\\\Rightarrow \Delta H=\dfrac{R\ln\dfrac{P}{P_0}}{\dfrac{1}{T_0}-\dfrac{1}{T_B}}\\\Rightarrow \Delta H=\dfrac{8.314\times \ln\left(\frac{100}{760}\right)}{\frac{1}{80.1+273.15}-\frac{1}{26.1+273.15}}\\\Rightarrow \Delta H=33008.99\ \text{J/kg}[/tex]
The heat of vaporization of benzene is 33009 J/kg.
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Which number represetns a coefficient?
2
3
4
7
A chemist prepares a solution of silver(II) oxide by measuring out 0.0013 of silver(II) oxide into a 100 mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's silver(II) oxide solution. Be sure your answer has the correct number of significant digits.
Answer:
1.3x10⁻⁸ mol/L
Explanation:
0.0013μmol, Calculate concentration in mol/L
To obtain concentration in mol/L we need to convert the μmoles to moles and mL to liters:
Moles silver(II) oxide:
0.0013μmol × (1mol / 1x10⁶μmol) = 1.3x10⁻⁹ moles
Liters solution:
100mL * (1L / 1000mL) = 0.1L
That means concentration in mol/L is:
1.3x10⁻⁹ moles / 0.1L =
1.3x10⁻⁸ mol/Lis carried out in a flow reactor where pure A is fed at a concentration of 4.0 mol/dm3. If the equilibrium conversion is found to be 60%, (a) What is the equilibrium constant, KC if the reaction is a gas phase reaction? (Ans.: Gas: KC = 0.328 dm3/mol)
Answer:
0.328 mol/dm³
Explanation:
We have
I started this calculation from Rate's law.
Remember equilibrium constant has been given to be 60%
Our interest is Kc, that is the equilibrium constant.
Ca = 4(1-0.6)/1+(-0.5*0.6)
= 4-2.4/1-0.3
= 1.6/0.7
= 2.2857
Cb = 4x0.6/2(1+(-0.5*0.6))
= 2.4/2(0.7)
= 2.4/1.4
= 1.7143
Kc = Cb/Ca²
= 1.7143/2.2857²
= 1.7143/5.2244
= 0.328 mol/dm³
I have added an attachment showing earlier stages to the final answer
Please help!!
This is a big part of my grade -----
Will make you brainliest******
Explanation:
U need to draw the graph first and make a line at 17 pennies, where the line of 17 pennies and your graph meet is the mass of it(at y axis)
The following are placed in a beaker weighing 39.457 g:
2.689 g of NaCl, 1.26 g of sand and 5.0 g water
What is the final mass of the beaker?
Answer:
48.4 g
Explanation:
The amount of force that is exerted on a balloon by the gas inside the balloon is.
O A) temperature
OB) prlessure
O C) volume
O D) heat
Answer:
pressure
Explanation:
pressure is the amount of force exerted on an area. when you blow up the balloon you're filling it with gas particles. the gas particles move freely within the balloon and may collide with one another exerting pressure on the inside of the balloon.
The pressure of the gas is the amount of force that is exerted on a balloon by the gas inside the balloon. Therefore, option B is correct.
What is pressure?Pressure can be described as the force applied perpendicular to the surface of a body per unit area. Pressure can be defined as a standard mechanical quantity and is derived from a unit of force divided by a unit of area.
The SI unit of measurement of pressure, the pascal (Pa) or Newton per square meter (N/m²). Pressure can be defined as the amount of force exerted perpendicular to the surface per unit area.
Mathematically, the pressure exerted by force can be calculated as:
[tex]{\displaystyle p={\frac {F}{A}}}[/tex]
where, p is the pressure, F is the magnitude of the normal force, and A is the area of the surface.
Therefore, the amount of force that is exerted on the balloon by the gas inside the balloon is equal to pressure.
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What is the Kc equilibrium-constant expression for the following equilibrium? S8(s) + 24F2(g) 8SF6(g)
Answer:
[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]
Explanation:
Hello.
In this case, for the undergoing chemical reaction:
[tex]S_8(s) + 24F_2(g) \rightleftharpoons 8SF_6(g)[/tex]
We consider the law of mass action in order to write the equilibrium expression yet we do not include S8 as it is solid and make sure we power each gaseous species to its corresponding stoichiometric coeffient (24 for F2 and 8 for SF6), thus we obtain:
[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]
Best regards!
How many atoms are in 10 g of He
Answer:
6.7
⋅
10
23
atoms of H
Explanation:
The empirical formula of CBr2 has a molar mass of 515.46 g/mol. What is the molecular formula of this
compound
Answer:
C3Br6
Explanation:
C= (1 X 12.011) = 12.011
Br= (2 X 79.904)= 159.808
159.808+12.011 = 171.819
515.46 divided by 171.819 = 3.00
so you mulitpy CBr2 by 3 which gives you C3Br6
How many moles are in 141.16 grams of F?
Use two digits past the decimal for all values
Answer: 2681.81
Explanation:
HELPP
describe what potassium would do to be more stable
Answer:
Explanation:
Its a elemental potassium is soft ,white in colour and has one more electron than argon,an element that we know is extremely stable ... Potassium extra electron is easily lost to form the much more stable cation, K+
Determine the value of the equilibrium constant (report your answer to three significant figures) for the following reaction if an equilibrium mixture contains 0.010 mol of solid PbBr2, and is 0.0100 M in Pb2+ ions and 0.0250 M in Br1- ions. Use the notation 4.31e-5 to indicate a number such as 4.31 x 10-5.
Answer:
6.25e-6 is the value of the equilibrium constant
Explanation:
we have this equation
[tex]PbBr(s) ----- Pb^{2+}(aq) + 2Br(aq)[/tex]
When at a state of equilibrium,
we have the concentration of Pb^2+ to be 0.01
we have the concentration of Br^- to be 0.025
the equilibrium constant concentration of both pure solids and liquid s are said to be equal to 1
[PbBR2] = 1
such tht
Keq = [Pb^2+] x [Br-]^2
we already know the values of these from the above.
0.01x0.025^2
= 0.01 x 0.000625
= 0.00000625
= 6.25 x 10^-6
= 6.25e^-6
How is matter divided?
True or False: The exact location of an electron can be measured thanks to
modern science.
Answer:
false you can not get a exact location of electrons from just modern science
Which of the following evidence supports the theory of plate tectonics
Answer:
seafloor spreading
Explanation:
i took the test
the half life of I -137 is 8.07 days. if 24 grams are left after 40.35 days, how many grams were in the original sample?
Answer:
768g
Explanation:
We can use to formula [tex]N(A) = N_0(\frac{1}{2})^\frac{t}{t_{1/2}}[/tex] . Here, N(A) is the final amount. N0 is the initial amount. t is the time elapsed, and [tex]t_{1/2}[/tex] is the half life. Plugging in, we get the answer above.
The half life of I -137 is 8.07 days. if 24 grams are left after 40.35 days, 800 gram were in the original sample.
What is half life?The half-life (symbol t12) is the amount of time it takes for a volume (of material) to be reduced to half of its original value. In nuclear physics, the phrase is typically used to indicate how rapidly unstable atoms experience radioactive decay or even how long stable nuclei survive.
The phrase is sometimes used more broadly to describe any form of exponential (or, in rare cases, non-exponential) decay. The biological ½ of medications and other compounds in the human body, for example, is referred to in the medical sciences. In exponential growth, the inverse of half-life is doubling time.
ln P = kt + C
P = amount of I-137 at time t
C = constant
k = 1/time
t = time
1st condition:
P = Po, t = 0 days
2nd condition: (half-life)
P = 0.5Po, t = 8.07 days
3rd condition:
P = 25 grams, t = 40.35 days
Po = 800 grams
mass of I-137 = 800 gram
Therefore, 800 gram were in the original sample.
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Calculate the pH of a solution containing a caffeine concentration of 455 mg/L . Express your answer to one decimal place.
Answer:
Explanation:
Caffeine is a weak base with pKb = 10.4
Kb = 10⁻¹⁰°⁴ = 3.98 x 10⁻¹¹
molecular weight of caffeine = 194.2
455 x 10⁻³ g / L = 455 x 10⁻³ / 194.2 moles / L
concentration of given solution a = 2.343 x 10⁻³ M
Let the caffeine be represented by B .
B + H₂O = BH + OH⁻
a - x x x
x² / ( a - x ) = Kb
x² / ( a - x ) = 3.98 x 10⁻¹¹
x is far less than a so a -x is almost equal to a
x² = 3.98 x 10⁻¹¹ x 2.343 x 10⁻³ = 9.32 x 10⁻¹⁴
x = 3.05 x 10⁻⁷
[ OH⁻ ] = 3.05 x 10⁻⁷
pOH = - log ( 3.05 x 10⁻⁷ )
= 7 - log 3.05
= 7 - 0.484 = 6.5
pH = 14 - 6.5 = 7.5
The pH of 455 mg/L of caffeine is 7.5
Using the formula;
Mass concentration = molar concentration × molar mass
Molar mass of caffeine = 194 g/mol
Mass concentration of caffeine = 455 mg/L
Molar concentration = Mass concentration/molar mass
Molar concentration = 455 × 10^-3g/L/194 g/mol
= 0.00235 M
Let Caffeine by depicted by the general formula BH
We can now set up the ICE table as follows;
:B + H2O ⇄ BH + OH^-
I 0.00235 0 0
C - x +x +x
E 0.00235 - x x x
Note that water is present in large excess
Again; pKb of caffeine =10.4
Kb = Antilog[-pKb]
Kb = Antilog [-10.4]
Kb = 3.98 × 10^-11
Kb = [BH] [OH^-]/[:B]
3.98 × 10^-11 = [x] [x]/[ 0.00235 - x ]
3.98 × 10^-11 [ 0.00235 - x ] = [x] [x]
9.4 × 10^-14 - 3.98 × 10^-11x = x^2
x^2 + 3.98 × 10^-11x - 9.4 × 10^-14 = 0
x = 3.1 × 10^-7 M
Recall [BH] = [OH^-] = 3.1 × 10^-7 M
Now;
pOH = - log [OH^-]
pOH = log [3.1 × 10^-7 M]
pOH = 6.5
But;
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 6.5
pH = 7.5
The pH of 455 mg/L of caffeine is 7.5
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Missing parts
Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4. Calculate the pH of a solution containing a caffeine concentration of 455 mg/L.
Calculate the theoretical density (in g/cm3) of copper (Cu), given that it has the FCC structure. The atomic weight of Cu is 63.55 g/mol, and its atomic radius R is 0.1278 nm.
Answer:
8.937g/cm³
Explanation:
To answer this question we need to know that, in 1 unit FCC cell you have:
Edge length = √8 * R
Volume = 8√8 * R³
And there are 4 atoms per unit cell
Mass of 4 atoms in g:
4 atom * (1mol / 6.022x10²³atom) * (63.55g / mol) = 4.221x10⁻²²g
Volume in cm³:
0.1278nm * (1x10⁻⁷cm / 1nm) = 1.278x10⁻⁸cm
Volume = 8√8 * (1.278x10⁻⁸cm)³
Volume = 4.723x10⁻²³cm³
And density is:
4.221x10⁻²²g / 4.723x10⁻²³cm³ =
8.937g/cm³