. Give the symbols of TWO elements that are in the same group.

Calculate the energy required to melt 21 g of ice at 0 oC.

vladimir1956 [14]

We are given:

Mass of ice = 21 grams

The ice is already at 0°c, the temperature at which it melts to form water

Molar heat of fusion of Ice = 6.02 kJ/mol

Finding the energy required:

Number of moles of Ice:

Molar mass of water = 18 g/mol

Number of moles = given mass/ molar mass

Number of moles = 21 / 18 = 7/6 moles

Energy required to melt the given amount of ice:

Energy = number of moles * molar heat of fusion

Energy = (7/6) * (6.02)

Energy = 7.02 kJ OR 7020 joules

7 0

3 years ago

Consider the reaction below.

Law Incorporation [45]

Answer:

First choice: 0.042

Explanation:

Given decomposition reaction:

1PCl₅ (g) ⇄ 1PCl₃ + 1Cl₂(g)

Equilibrium constant:

$K_{eq}=\frac{[PCl_3]^1[Cl_2]^1}{[PCl_5]^1}$

Stoichiometric coefficients and powers equal to 1 are not usually shown as they are understood, but I included them in order to shwow you how they intervene in the equilibrium expressions: each concentration is raised to a power equal to the respective stoichiometric coefficient in the equilibrium equation.

So, your calculations are:

$K_{eq}=\frac{(0.020M)(0.020M)}{0.0095M}=0.042M$

6 0

3 years ago

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

50

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$