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FrozenT [24]
4 years ago
11

Classify each of the following binary compounds by the oxidation number of hydrogen. Match each item with appropriate category.

items: H2O, HCl, MgH2, NH3, NaH, CaH2,H2,CH4 cateogories -1, 1+, 0
Chemistry
1 answer:
lozanna [386]4 years ago
6 0

Answer:

1. Category +1: H^{+1}_{2}O^{-2}, H^{+1}Cl^{-1}, N^{-3}H^{+1}_{3}, Na^{-1}H^{+1} ,C^{-4}H^{+1}_{4}

2. Category -1: Mg^{+2}H^{-1}_{2}, Ca^{+2}H^{-1}_{2}

3. Category 0: H^{0}_{2}

Explanation:

1. Category +1:

The more common number of oxidation of the oxygen is +1.

H^{+1}_{2}O^{-2}

H^{+1}Cl^{-1}

N^{-3}H^{+1}_{3}

Na^{-1}H^{+1}

C^{-4}H^{+1}_{4}

2. Category -1:

The hydrogen takes the number of oxydation -1 in the hydrides.

Mg^{+2}H^{-1}_{2}

Ca^{+2}H^{-1}_{2}

3. Category 0:

The number of oxydation of an isolated element is always 0.

H^{0}_{2}

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Explanation:

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A sample of gas occupies a volume of 67.5 mL . As it expands, it does 131.0 J of work on its surroundings at a constant pressure
baherus [9]

Answer:

The final volume V2=1.3175L

Explanation:

between work ( w), pressure ( P ) and volume ( V ) is the following:

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CHECK THE ATTACHMENT FOR DETAILED EXPLATION

4 0
4 years ago
If a compound has a composition of 82% nitrogen and 18% hydrogen, what is the empirical formula for this compound
Damm [24]

Answer: The empirical formula for the given compound is NH_3

Explanation : Given,

Percentage of H = 18 %

Percentage of N = 82 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of H = 18 g

Mass of N = 82 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{18g}{1g/mole}=18moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{82g}{14g/mole}=5.8moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 5.8 moles.

For Hydrogen  = \frac{18}{5.8}=3.10\approx 3

For Nitrogen = \frac{5.8}{5.8}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of H : N = 3 : 1

Hence, the empirical formula for the given compound is NH_3

3 0
3 years ago
A sample of an unknown gas takes 222 s to diffuse through a porous plug at a given temperature. At the same temperature, N2(g) t
gulaghasi [49]

Answer:

\large \boxed{\text{45.1 g/mol}}

Explanation:

Graham’s Law applies to the diffusion of gases:

The rate of diffusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

r \propto \dfrac{1}{\sqrt{M}}

If you have two gases, the ratio of their rates of diffusion is

\dfrac{r_{2}}{r_{1}} = \sqrt{\dfrac{M_{1}}{M_{2}}}

The time for diffusion is inversely proportional to the rate.

\dfrac{t_{2}}{t_{1}} = \sqrt{\dfrac{M_{2}}{M_{1}}}

Data:

t₂ = 222 s

 t₁ = 175 s

M₁ = 28.01

Calculation :

\begin{array}{rcl}\dfrac{222}{175} & = & \sqrt{\dfrac{M_{2}}{28.01}}\\\\1.269 & = & \sqrt{\dfrac{M_{2}}{28.01}}\\\\1.609 & = & \dfrac{M_{2}}{28.01}\\\\M_{2} & = & 1.609 \times 28.01\\ & = & \textbf{45.1 g/mol}\\\end{array}\\\text{The molar mass of the unknown gas is $\large \boxed{\textbf{45.1 g/mol}}$}

3 0
3 years ago
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