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AfilCa [17]
3 years ago
11

Differences in the moons and suns pull on different sides of earth cause _______

Chemistry
1 answer:
Aleksandr [31]3 years ago
5 0

Answer:

tides are caused mainly by the differences in how much gravity from the moon and the sun pulls on different parts of earth

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At a given temperature, 3.12 atm of H2 and 5.52 atm of I2 are mixed and allowed to come to equilibrium. The equilibrium pressure
expeople1 [14]

Answer: 0.596

Explanation:

For this problem, we want to find K_{p}. To do so, we will need to use the ICE chart. The I in ICE is initial quantity. In this case, it is the initial pressure. Pressure is in atm. The C in ICE is change in each quantity. The E is equilibrium.

          H₂(g) + I₂(g) ⇄ 2HI(g)

I          3.12     5.52         0

C      -0.869  -0.869   +1.738

E       2.251    2.251     1.738

<u>For the steps below, refer to the ICE chart above.</u>

1. Since we were given the initial of H₂, I₂ and equilibrium of 2HI, we can fill those into the chart.

2. Since we were not given the initial for 2HI, we will put 0 in their place.

3. For the change, we need to add pressure to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x. We know that the equilibrium of 2HI is 1.738, we know the change is 0.869 because 1.738/2=0.869. Since there are 2 moles of HI, we must divide the equilibrium by 2 to find x, so that we can fill that into the reactants side.

4. With the equilibrium values, we can find the equilibrium pressure. It is products over reactants. We use the values of E.

K_{p} =\frac{[HI]^2}{[H_{2}][I_{2} ] }

The [HI]² comes from 2HI. We more the moles to the exponent when we are calculating the equilibrium pressure.

K_{p} =\frac{(1.738)^2}{(2.251)(2.251)} =0.596

We typically don't put units because K is unitless, but know that the units is atm throughout the entire problem.

5 0
3 years ago
Please help me, I’m stuck on this question
zubka84 [21]

Answer:

\large \boxed{\text{0.0820 L$\cdot$atm$\cdot$K$^{-1}$mol}^{-1 }}

Explanation:

To solve this problem, we can use the Ideal Gas Law:

pV = nRT

Data:

p = 3.00 atm

V = 17.4 L

n = 2.00 mol

T = 45 °C

Calculations:

1. Convert the temperature to kelvins

T = (45 + 273.15) K = 318.15 K

2. Calculate the value of R

\begin{array}{rcl}pV & = & nRT\\\text{3.00 atm} \times \text{17.4 L} & = & \text{2.00 mol} \times R \times \text{318.15 K}\\\text{52.2 L$\cdot$atm} & = & 636.30R \text{ K$\cdot$mol}\\R & = & \dfrac{\text{52.2 L$\cdot$atm}}{636.30 \text{ K$\cdot$mol}}\\\\ & = & \textbf{0.0820 L$\cdot$atm$\cdot$K$^{-1}$mol}^{-1} \\\end{array}\\\text{The value of the gas constant R is $\large \boxed{\textbf{0.0820 L$\cdot$atm$\cdot$K$^{-1}$mol}^{-1 }}$}

7 0
3 years ago
1. 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g) a. Determine the volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g
QveST [7]

Answer:

a. 167 mL b. 39.27 %

Explanation:

a. From the chemical equation. 2 mole of Al reacts with 3 mole H₂SO₄ to produce 1 mol  Al₂(SO₄)₃.

Now, we calculate the number of moles of Al in 45.0 g Al.

We know number of moles, n = m/M where m = mass of Al = 45.0 g and M = molar mass of Al = 26.98 g/mol.

So n = 45.0 g/26.98 g/mol = 1.668 mol

Since 2 mole of Al reacts with 3 mole H₂SO₄, then 1.668 mole of Al reacts with x mole H₂SO₄. So, x = 3 × 1.668/2 mol = 2.5 mol

So, we have 2.5 mol H₂SO₄.

Now number of moles of H₂SO₄, n = CV where C = concentration of H₂SO₄ = 15.0 M = 15.0 mol/L and V = volume of H₂SO₄.

V = n/C

= 2.5 mol/15.0 mol/L

= 0.167 L

= 167 mL of 15.0 M H₂SO₄ reacts with 45.0 g Al to produce aluminum sulfate.

b. From the chemical reaction, 2 mol Al produces 1 mol Al₂(SO₄)₃

Therefore 1.668 mol Al will produce x mol  Al₂(SO₄)₃. So, x = 1 mol × 1.668 mol/2 mol = 0.834 mol

So, we need to find the mass of 0.834 mol  Al₂(SO₄)₃. Now molar mass  Al₂(SO₄)₃ = 2 × 26.98 g/mol + 3 × 32 g/mol + 4 × 3 × 16 g/mol = 53.96 g/mol + 96 g/mol + 192 g/mol = 341.96 g/mol.

Also number of moles of  Al₂(SO₄)₃, n = mass of  Al₂(SO₄)₃,m/molar mass  Al₂(SO₄)₃, M

n =m/M

So, m = nM = 0.834 mol × 341.96 g/mol = 285.2 g

% yield = Actual yield/theoretical yield × 100 %

Actual yield = 112 g, /theoretical yield = 285.2 g

So, % yield = 112 g/285.2 g × 100 %

= 0.3927 × 100 %

=  39.27 %

8 0
3 years ago
PHET Simulator
AlekseyPX

Answer:

there is 1 leftover because your adding to 1... I think

Explanation:

3 0
3 years ago
Which of the following is true for all types of fronts ( warm, cold, stationary, and occluded)
Nadusha1986 [10]
D). A warm air mass is in contact with a cold air mass. i hope its right PLEASE MARK ME BRAINLEST
6 0
4 years ago
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