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3 years ago

For a reaction system at equlibrium, le chateliers principle can be used to predict the

1 answer:

6 0

According to Le Chatelier's principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. ... By Le Chatelier's principle, we can predict that the amount of methanol will increase, thereby decreasing the total change in CO.

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Which profession is likely to use a radiation shield?

pantera1 [17]

Which profession is likely to use a radiation shield?
A.)dentist
B.)astronomer
C.)mathematician
D.)geologist

The answer is B: Astronomer

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3 years ago

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What determines the average kinetic energy of the particles in a gas?

jolli1 [7]

Answer:

The temperature.

Explanation:

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2 years ago

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Currents at the surface of the ocean are

lakkis [162]

Answer:

D. The mixing of warm and cold water

Explanation:

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3 years ago

A 33.0 mL sample of 1.15 M KBr and a 59.0 mL sample of 0.660 M KBr are mixed. The solution is then heated to evaporate water unt

Oksi-84 [34.3K]

Answer:

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

Explanation:

Step 1: Data given

Sample 1: The 1.15 M sample has a volume of 33.O mL

Sample 2: The 0.660 M sample has a volume of 59.0 mL

Molar mass of KBr = 119 g/mol

Molar mass of AgNO3 = 169.87 g/mol

Step 2: Calculate number of moles for both samples

Number of moles = Molarity * Volume

Sample 1: 1.15 M * 33 *10^-3 L = 0.03795 moles

Sample 2: 0.660 M *59*10^-3 L = 0.03894 moles

Total mol KBr = 0.03795 + 0.03894 = 0.07689 moles

Step 3: Calculate total mass

mass = Number of moles * Molar mass

mass = 0.07689 moles * 119 g/moles = 9.15 grams ( in 55mL)

Step 4: Calculate moles of AgBr

AgNO3 reacts with KBr

KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)

1 mole of KBr consumed, needs 1 mole of AgNO3 to produce 1 mole of AgBr and 1 mole of KNO3

So 0.07689 moles of KBr wll need 0.07689 moles of AgNO3

Step 5: Calculate mass of silver nitrate

mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3

mass of AgNO3 = 0.07689 moles * 169.87 g/mol = 13.06 grams

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

8 0

3 years ago

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is:

$s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}$

$s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}$

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0

2 years ago