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SashulF [63]
2 years ago
7

CO(g) + 3 H2(g) → CH4(k) + H2O(g) + heat

Chemistry
1 answer:
weqwewe [10]2 years ago
8 0

Explanation:

AH is negative because the reaction gives out heat (exothermic).

AS is negative because the total entropy of the reactants is greater than that of the products (there are 4 moles of gas on reactant side compared to only 1 mole on the product side). Thus, the reactants are said to be "messier" than the products.

AG is negative at lower temperatures because at low temperatures, -TAS is more negative, resulting in the negative sign of AG.

Hope this helps.

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3d10 4s2 4p5. I think
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A gas occupies a volume of 30.0L, a temperature of 25°C and a pressure of 0.600atm. What will be the volume of the gas at STP?​
Shalnov [3]

Answer:

=16.49 L

Explanation:

Using the equation

P1= 0.6atm V1= 30L, T1= 25+273= 298K, P2= 1atm, V2=? T2= 273

P1V1/T1= P2V2/T2

0.6×30/298= 1×V2/273

V2=16.49L

5 0
3 years ago
Definition: A compound composed of two elements that share valence electrons
Evgen [1.6K]

Answer:

Explanation:

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3 0
3 years ago
Indicate which of the following criteria are important for the selection of a buffer to use in an in vitro biochemical reaction.
JulsSmile [24]

Answer:

The correct answer is: d. The pKa of the chosen buffer should be close to the optimal pH for the biochemical reaction.

Explanation:

The buffer resist or maintain the change in pH in case of Acid or basic addition to the solution. The buffer capacity should be within one or two pH units when compared to the optimal pH.

Thus it is important to select a buffer with pKa close to the optimum pH of the reaction because the ability for the buffer to maintain the pH is is great at the pH close to pKa.

7 0
2 years ago
a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re
Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

Given : 59.4 g of H_2SO_4 in 100 g of solution  

moles of H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61

Volume of solution =\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml

Now put all the given values in the formula of molality, we get

Molality=\frac{0.61\times 1000}{54.6ml}=11.2M

To calculate the volume of acid, we use the equation given by neutralisation reaction:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of stock acid which is H_2SO_4

M_2\text{ and }V_2 are the molarity and volume of dilute acid which is H_2SO_4

We are given:

M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL

Putting values in above equation, we get:

11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

4 0
3 years ago
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