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Chemistry

1 answer:

Anit [1.1K]3 years ago

5 0

Answer:

B is correct answer because The enthalpy is negative so this means the reaction produces heat and the reaction is exothermic

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Acid + base —> salt + water What type of reaction is this?

erastova [34]

$\sf{Acid}+\bf{Base}\longrightarrow \tt{Salt}+\bold{Water }$

These type of reaction is called as Neutralization Reaction.

8 0

3 years ago

Nitrogen and hydrogen gases are combined at high temperatures and pressures to produce ammonia, NH3. If 101.7 g of N2 are reacte

Lerok [7]

Answer:

7.26 moles of NH₃ are formed in this reaction

Explanation:

This is about the reaction for the production of ammonia

1 mol of nitrogen gas reacts to 3 moles of hydrogen in order to produce 2 moles of ammonia.

The equation is: N₂ + 3H₂ → 2NH₃

In the question, we were informed that the excess is the H₂ so the N₂ is limiting reagent. We determine the moles, that has reacted:

101.7 g / 28 g/mol = 3.63 moles

So, If 1 mol of nitrogen gas can produce 2 moles of ammonia

3.63 moles of N₂ must produce ( 2 . 3.63) / 1 = 7.26 moles of NH₃

6 0

4 years ago

Read 2 more answers

A sample of water has a mass of 193 g. Salt is added to the water, and the final solution weighs 238 g. How much salt was added?

icang [17]

Answer:

45 g of salt was added

Explanation:

238g-193g=45g

3 0

3 years ago

Calculate the molecular weight of a dibasic acid.0.56gm of which is required 250ml of N/20 sodium hydroxide solution for neutral

Alekssandra [29.7K]

Answer: The molecular weight of the dibasic acid is 89.6 g/mol

Explanation:

Normality is defined as the amount of solute expressed in the number of gram equivalents present per liter of solution. The units of normality are eq/L. The formula used to calculate normality:

$\text{Normality}=\frac{\text{Given mass of solute}\times 1000}{\text{Equivalent mass of solute}\times \text{Volume of solution (mL)}}$ ....(1)

We are given:

Normality of solution = $\frac{1}{20}=0.05N$

Given mass of solute = 0.56 g

Volume of solution = 250 mL

Putting values in equation 1, we get:

$0.05=\frac{0.56\times 1000}{\text{Equivalent mass of solute}\times 250}\\\\\text{Equivalent mass of solute}=\frac{0.56\times 1000}{0.05\times 250}=44.8g/eq$