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3 years ago

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

Chemistry

1 answer:

weqwewe [10]3 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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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?

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=16.49 L

Explanation:

Using the equation

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

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3 years ago

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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.

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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

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3 years ago