All categories

3 years ago

Examine the reaction. NH4OH(aq) →H2O(l) + NH3(g)

Chemistry

1 answer:

shusha [124]3 years ago

7 0

Answer: A. 1,1,1

Explanation:

The coefficients that will balance the equation; NH4OH(aq) →H2O(l) + NH3(g), is 1, 1, 1, because it proves the total number of atoms of each element on the LHS and RHS of the equation are equal, hence balanced.

LHS RHS

N = 1 1

H = 5 5

O = 1 1

You might be interested in

Review energy changes by identifying the correct statements about each reaction.

Grace [21]

Answer: The reaction is exothermic

The $\Delta H$ value is negative.

Heat is a product.

Explanation:

$2H_2(g)+O_2(g)\rightarrow 2H_2O(l)+136.6kcal$

Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and $\Delta H$ for the reaction comes out to be positive.

Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and $\Delta H$ for the reaction comes out to be negative.

The substances which are written on the left side of the arrow are reactants and the substances which are written on the right side of the arrow are products. Thus heat is a product.

5 0

3 years ago

Which compound is a hydrocarbon?

deff fn [24]

Answer:

C2H6 OR C6H12O6 THE ANSWER

3 0

2 years ago

Read 2 more answers

Is sugar made of tiny particles

netineya [11]

Sugar and water are made with tiny particles. They are both made from molecules and atoms.

6 0

3 years ago

Given the equation:

My name is Ann [436]

Answer:

this isnt even a question...

Explanation: what the heck

6 0

3 years ago

he heat of fusion of tetrahydrofuran is . Calculate the change in entropy when of tetrahydrofuran melts at . Be sure your answer

lana [24]

Answer:

$\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}$

Explanation:

Hello.

In this case, given the heat of fusion of THF to be 8.5 kJ/mol and freezing at -108.5 °C, for the required mass of 5.9 g, we can compute the entropy as:

$\Delta S=\frac{n*\Delta H}{T}$

Whereas n accounts for the moles which are computed below:

$n=5.9g*\frac{1mol}{72g} =0.082mol$

Thus, the entropy turns out:

$\Delta S=\frac{0.0819mol*8.5 kJ/mol}{(-108.5+273.15)K}\\\\\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}$

Best regards.

3 0

3 years ago