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Nadusha1986 [10]

3 years ago

13

This chemical reaction is unbalanced: _Pb(NO₃)₂(aq) + _Li₂SO₄(aq) → PbSO₄(s) + LiNO₃(aq) Which coefficient should ap

pear in front of LiNO₃ in the balanced equation?
1
2
3
4

1 answer:

kakasveta [241]3 years ago

4 0

Answer: 2

Explanation:

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A scientist finds that one volume of an unknown gas X combines with two volumes of another gas Y to form two volumes of a new ga

MariettaO [177]

Answer:

Gas X

Explanation:

The given reaction can be written in the form of chemical equation as shown below as:

$X+2Y\rightarrow 2XY_2$

According to law of conservation of mass, the moles of each substance in the reaction must be equal on both reactant and product side.

Also, the question asks for the gas which is diatomic. $XY_2$ cannot be diatomic as the formula contains 3 atoms.

Between gas X and gas Y , <u>X has to be diatomic for the reaction to balance </u>as:

$X_2+2Y\rightarrow 2XY_2$

3 0

3 years ago

A sample of gas X occupies 10 mº at a pressure of 120 kPa.

Mashutka [201]

Answer:

The new pressure of the gas comes out to be 400 KPa.

Explanation:

Initial volume of gas = V = $10\textrm{ m}^{3}$

Initial pressure of gas = P = 120 KPa

Final volume of gas = V' = $3\textrm{ m}^{3}$

Assuming temperature to be kept constant.

Assuming final pressure of the gas to be P' KPa

$PV = P'V' \\120\textrm{ KPa}\times 10\textrm{ m}^{3} = \textrm{P'}\times 3\textrm{ m}^{3} \\\textrm{P'} = 400\textrm{ KPa}$

New pressure of gas = 400 KPa

5 0

2 years ago

If some salt were accidently spilled as it was transferred from the balance to the cup would your calculated enthalpy of solutio

morpeh [17]

Answer:

It would be to low

Explanation:

7 0

2 years ago

has a standard free‑energy change of − 3.59 kJ / mol at 25 °C. What are the concentrations of A , B , and C at equilibrium if, a

Mamont248 [21]

Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.

Explanation:

The reaction equation is as follows.

$A + B \rightarrow C$

Initial : 0.3 0.4 0

Change: -x -x x

Equilbm: (0.3 - x) (0.4 - x) x

We know that, relation between standard free energy and equilibrium constant is as follows.

$\Delta G = -RT ln K$

Putting the given values into the above formula as follows.

$\Delta G = -RT ln K$

$-3.59 kJ/mol = -8.314 \times 10^{-3} kJ/mol K ln (\frac{x}{(0.3 - x)(0.4 - x)})$

x = 0.1417

Hence, at equilibrium

[A] = 0.3 - 0.1417

= 0.1583 M

[B] = 0.4 - 0.1417

= 0.2583 M

[C] = 0.1417 M

5 0

2 years ago

Calculate ΔH for the reaction: C(graphite) + 2H 2(g) + 1/2 O 2(g) => CH 3OH(l) Using the following information: C(graphite) +

Alika [10]

Answer:

$\Delta H$ for the given reaction is -238.7 kJ

Explanation:

The given reaction can be written as summation of three elementary steps such as:

$C(graphite)+O_{2}(g)\rightarrow CO_{2}(g)$ $\Delta H_{1}= -393.5 kJ$

$2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)$ $\Delta H_{2}= (2\times -285.8)kJ$

$CO_{2}(g)+2H_{2}O(l)\rightarrow CH_{3}OH(l)+\frac{3}{2}O_{2}(g)$ $\Delta H_{3}= 726.4 kJ$

---------------------------------------------------------------------------------------------------

$C(graphite)+2H_{2}(g)+\frac{1}{2}O_{2}(g)\rightarrow CH_{3}OH(l)$

$\Delta H=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}=-238.7 kJ$

4 0

3 years ago