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3 years ago

Explain why during the month of June the days are longer than nights at the North Pole

Chemistry

1 answer:

Oksanka [162]3 years ago

5 0

Well, the region is facing the sun, and hardly points away for half of a year.

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Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O

Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

I hope it helps!

5 0

3 years ago

Which of the following locations would experience the highest rate of sublimation?

g100num [7]

I think its the arctic ice caps

6 0

3 years ago

Read 2 more answers

A 256 mL sample of HCl gas is in a flask where it exerts a force (pressure) of 67.5 mmHg. What is the pressure of the gas if it

Effectus [21]

Answer:

The pressure in the new flask would be $128\; \rm mmHg$ if the $\rm HCl$ here acts like an ideal gas.

Explanation:

Assume that the $\rm HCl$ sample here acts like an ideal gas. By Boyle's Law, the pressure $P$ of the gas should be inversely proportional to its volume $V$ .

For example, let the initial volume and pressure of the sample be $V_1$ and $P_1$ . The new volume $V_2$ and pressure $P_2$ of this sample shall satisfy the equation: $P_1 \cdot V_1 =P_2 \cdot V_2$ .

In this question,

The initial volume of the gas is $V_1= 256\; \rm mL$ .
The initial pressure of the gas is $P_1 = 67.5\; \rm mmHg$ .
The new volume of the gas is $V_2 = 135\; \rm mL$ .

The goal is to find the new pressure of this gas, $P_2$ .

Assume that this sample is indeed an ideal gas. Then the equation $P_1 \cdot V_1 =P_2 \cdot V_2$ should still hold. Rearrange the equation to separate the unknown, $P_2$ . Note: make sure that the units for $V_1$ and $V_2$ are the same before evaluating. That way, the unit of

$\begin{aligned} & P_2\\ &= \frac{P_1 \cdot V_1}{V_2} \\ &= \frac{256\; \rm mL \times 67.5\; \rm mmHg}{135\; \rm mL} \\ & \approx 128\; \rm mmHg\end{aligned}$ .