Answer:
C.) balance in number of electrons and protons.
Explanation:
Its the right answer i swear!!!
Great amounts of atomic energy are released when a _______reaction occurs.
Great amounts of atomic energy are released when a chemical reaction occurs. The process can be an exothermic reaction or endothermic reaction depending on the substances involved in the reaction.
Answer:
For a: The mass percent of NaBr is 7.03 %
For b: The mass percent of KCl is 16.94 %
For c: The mass percent of toluene is 13.43 %
Explanation:
To calculate the mass percentage of solute in solution, we use the equation:
.......(1)
We are given:
Mass of NaBr (Solute) = 5.50 g
Mass of solution = 78.2 g
Putting values in equation 1, we get:
Hence, the mass percent of NaBr is 7.03 %
We are given:
Mass of KCl (Solute) = 31.0 g
Mass of water (solvent) = 152 g
Mass of solution = (31.0 + 152) g = 183 g
Putting values in equation 1, we get:
Hence, the mass percent of KCl is 16.94 %
We are given:
Mass of toluene (Solute) = 4.5 g
Mass of benzene (solvent) = 29 g
Mass of solution = (4.5 + 29) g = 33.5 g
Putting values in equation 1, we get:
Hence, the mass percent of toluene is 13.43 %
To calculate the percent by mass of the solute in each aqueous solution, divide the mass of the solute by the mass of the solution and multiply by 100%
To calculate the percent by mass of the solute in each aqueous solution, you'll need to use the formula:
Percent by mass = (mass of solute/mass of solution) x 100%
For example, in solution (a) with 5.50 g of NaBr in 78.2 g of solution, the mass of the solute is 5.50 g and the mass of the solution is 78.2 g. Plugging these values into the formula gives:
Percent by mass = (5.50 g / 78.2 g) x 100% = 7.03%
Similarly, you can calculate the percent by mass for solutions (b) and (c) using the same formula.
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Answer:
1.74845
Explanation:
We have the following reaction:
I2 + H2 => 2 HI
Now, the constant Kc, has the following formula:
Kc = [C] ^ c * [D] ^ d / [A] ^ a * [B] ^ b
In this case I2 is A, H2 is B and C is HI
We know that the values are:
H2 = 1 × 10 ^ -3 at 448 ° C
I2 = 2 × 10 ^ -3 at 448 ° C
HI = 1.87 × 10 ^ -3 at 448 ° C
Replacing:
Kc = [1.87 × 10 ^ -3] ^ 2 / {[2 × 10 ^ -3] ^ 1 * [1 × 10 ^ -3] ^ 1}
Kc = 1.87 ^ 2/2 * 1
Kc = 1.74845
Which means that at 448 ° C, Kc is equal to 1.74845
Answer:
Explanation:
[H2] = 10^-3
[I2] = 2*10^-3
[HI] = 0
in equilbiirum
[H2] = 10^-3 - x
[I2] = 2*10^-3 -x
[HI] = 0 + 2x
and we know
[HI] = 0 + 2x = 1.87*10^-3
x = ( 1.87*10^-3)/2 = 0.000935
then
[H2] = 10^-3 - 0.000935 = 0.000065
[I2] = 2*10^-3 -0.000935 = 0.001065
H₂ + I ⇄ 2 HI
Initially 1 × 10⁻³ 2 × 10⁻³
Change -9.35 × 10⁻⁴ -9.35 × 10⁻⁴ +1.87 × 10⁻³
At equil 6.5 × 10⁻⁵ 1.06 5 × 10⁻³ 1.87 × 10⁻³
HI increase by 1.87 × 10⁻³M
The breaking and making of chemical bonds can explain chemical reactions and energy flow. a. True b. False
True
The initial volume of the gas is 117L.
Volume is the amount of space the matter occupies.
PV/T = k
where P = pressure, V = volume, T = absolute temperature in K and k= constant.
P1V1 / T1 = P2V2 / T2
Where P1 and P2 = initial and final pressure in atm
T1 and T2 = Initial and final temperature in K
V1 and V2 = initial and final volume in L
Using combined gas law:
P1V1/T1 = P2V2/T2
Given:
P1 = 0.5 atm
T1 = 325K
P2 = 1.2 atm
V2 = 48L
T2 = 320K
Assuming that the number of moles are constant for both conditions:
(P1 x V1)/T1 = (P2 x V2)/T2
Substituting the given values,
(0.5 x V1)/325 = (1.2 x 48)/320K
Solving for V1
V1 = (1.2atm x 48L x 325K)/ (320Kx 0.5 atm)
V1 = 117L
Hence, the initial volume of the gas is 117L.
To learn more about combined gas law here
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