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zhenek [66]
3 years ago
8

Making methanol the element hydrogen is not abundant in nature, but it is a useful reagent in, for example, the potential synthe

sis of the liquid fuel methanol from gaseous carbon monoxide: $$2h2​(g)+co(g) ch3​oh(g) δh°f, kj/mol δs°f, j/mol·k h2(g) 0 130.6 co(g) -110.5 197.7 ch3oh(g) -238.7 126.8 1st attempt see hintsee periodic table this reaction is spontaneous at temperatures lower than:
Chemistry
1 answer:
stealth61 [152]3 years ago
5 0
ΔG⁰ = ΔH⁰ - TΔS
ΔH⁰ = Hf,(CH₃OH) - Hf,(CO) = -238.7 + 110.5 = -128.2 kJ/mol
ΔS = S(CH₃OH) - S(CO) - 2S(H₂) = 126.8 - 197.7 - 2 x 130.6 = -332.1 J/mol.K
So 
ΔG⁰ = - 128200 + 332.1 T
For the reaction to be spontaneous:
ΔG⁰ < 0
So: -128200 + 332.1 T < 0
332.1 T < 128200
T < 386.028 K
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Answer:

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Explanation:

3 0
3 years ago
How can you determine the specific heat of a metal using a calorimeter
zhannawk [14.2K]

Answer:

One can determine the specific heat of the metal through using the clarimeter, water, thermometer and using heat equations.

Explanation:

You can learn about heat effects and calorimetery through a simple experiment by boiling water and heating up the metal in it. Then, pour it into your calorimeter and the heat will flow from the metal to the water. The two equlibria will meet: the metal will loose heat into its surroundings (the water) and teh water will absorb the heat. The heat flow for the water is the same as it is for the metal, the only difference being is the negative sign indicating the loss of the heat of the metal.

In terms of theromdynamics, we can deteremine the heat flow for the metal becasue it would be equal to the mangnitued but opposite in direction. Thus, we can say that the specific heat of water qH2O = -qmetal.

4 0
3 years ago
Jamie is measuring the mass of one mole of carbon. She measures the mass 4 times, achieving the following measurements: 11.86 g,
azamat

This is an incomplete question, here is a complete question.

Distinguish between Accuracy and Precision Question Jamie is measuring the mass of one mole of carbon. She measures the mass 4 times, achieving the following measurements: 11.86 g, 11.87 g, 11.89 g, and 11.90 g. If the accepted value of one mole of carbon is 12.11 g, how can Jamie's measurements best be described?

Select the correct answer below:

Jamie's measurements are both precise and accurate,

Jamie's measurements are accurate but not precise.

Jamie's measurements are precise but not accurate,

Jamie's measurements are neither precise nor accurate,

Answer : The correct option is, Jamie's measurements are precise but not accurate.

Explanation :

Accuracy : It is defined as the closeness of a measured value to a standard or known value.

For Example: If the mass of a substance is 50 kg and one person weighed 48 kg and another person weighed 55 kg. Then, the weight measured by first person is more accurate.

Precision : It is defined as the closeness of two or more measurements to each other.

For Example: If you weigh a given substance five times and you get 1.8 kg each time. Then the measurement is said to be precise.

Level of precision is determined by the maximum number of decimal places.

As we are given that:

The measurements are, 11.86, 11.87, 11.89 and 11.90

The accepted value is, 12.11

The average of these values =  \frac{11.86+11.87+11.89+11.90}{4}=11.88

From this we conclude that, these measurements are close to each other that means they are precise. But the accepted value are not equal to average of four values that means they are not accurate.

Hence, the correct option is, Jamie's measurements are precise but not accurate.

7 0
4 years ago
Consider the hypothetical serine protease, which shows the specificity pockets. The S1 pocket has a glutamic acid in the bottom,
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Answer:

See explaination

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Please see attachment for the step by step solution of the given problem.

8 0
3 years ago
Help me please) attached the screen below. Thanks
Vitek1552 [10]

Answer:

1) d = 2.4 g/cm³

2) m = 25 g

3) v = 126.7 cm³

Explanation:

Given data:

Mass of material = 24 g

Volume of material = 10 cm³

Density of material = ?

Solution:

Formula:

d = m/v

by putting value,

d = 24 g / 10 cm³

d = 2.4 g/cm³

2) Given data:

Density of material = 5 g/cm³

Volume of material = 5 cm³

Mass of material = ?

Solution:

Formula:

d = m/v

5 g/cm³ = m / 5 cm³

m = 5 g/cm³×5 cm³

m = 25 g

3)Given data:

Density of material = 3 g/cm³

Mass of material = 380 g

Volume of material =  ?

Solution:

Formula:

d = m/v

3 g/cm³ = 380 g / v

v = 380 g /3 g/cm³

v = 126.7 cm³

7 0
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