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3 years ago

Which molecule contains the smallest number of hydrogen atoms? A) N2H4. B) Al(OH) 3. C) (NH4) 2SO4. D) NH4NO3.

Chemistry

2 answers:

Vesna [10]3 years ago

7 0

The molecule that contains the fewest number of Hydrogen atoms would be B. Al(OH)3. It only has 3 Hydrogen atoms.

Svetach [21]3 years ago

3 0

Answer: Option (B) is the correct answer.

Explanation:

In a molecule of $N_{2}H_{4}$ , there are 2 nitrogen atoms and 4 hydrogen atoms.

In a molecule of $Al(OH)_{3}$ , there are one aluminium atom, three oxygen atoms, and three hydrogen atoms.

In a molecule of $(NH_{4})_{2}SO_{4}$ , there are 2 nitrogen atoms, 8 hydrogen atoms, one sulfur atom and 4 oxygen atoms.

In a molecule of $NH_{4}NO_{3}$ , there are 2 nitrogen atoms, 4 hydrogen atoms and 3 oxygen atoms.

Therefore, we can conclude that out of the given options $Al(OH)_{3}$ is the molecule which contains the smallest number of hydrogen atoms.

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erastovalidia [21]

Answer:80KM is distance. 30KM north is displacement.

Explanation:

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2 years ago

5. calculate the percentage of water of crystallization in blue vitriol (copper sulphate pentahydrate), cuso4.5h2o

ki77a [65]

The percentage of water of crystallization in blue vitriol is 36.07%.

M(H₂O) = 2Ar(H) + Ar(O) x g/mol

M(H₂O) = 2 + 16 x g/mol

M(H₂O) = 18 g/mol; molar mass of water

M(CuSO₄·5H₂O) = Ar(Cu) + Ar(S) + 4Ar(O) + 5Mr(H₂O) x g/mol

M(CuSO₄·5H₂O) = 63.5 + 32 + 64 + 90 x g/mol

M(CuSO₄·5H₂O) = 249.5 g/mol; molar mass of copper sulphate pentahydrate

The percentage of water: 5M(H₂O) / M(CuSO₄·5H₂O) x 100%

The percentage of water: 90 g/mol / 249.5 g/mol x 100%

The percentage of water = 36.07%

More about blue vitriol: brainly.com/question/8895853

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5 0

1 year ago

Read 2 more answers

Three blocks of metal at the same temperature are placed on a hot stove. Their specific heat capacities are listed below.Rank th

JulijaS [17]

Answer:

Copper>Steel>Aluminium

Explanation:

Hello,

Since the heat capacity accounts for the required heat to increase by 1°C, 1 kg of the metal, copper is the one that has the lower heat capacity, it means that it requires the least amount of energy to warm up (increase its temperature), this could be substantiated via the mathematical definition of heat capacity:

$H=mC(T_2-T_1)$

Solving for $T_2$ :

$T_2=T_1+\frac{H}{mC}$

It means that the lower the heat capacity, the higher the final temperature.

Best regards.

7 0

3 years ago

Suppose you needed to measure exactly 15.0 mL of water for an

s344n2d4d5 [400]

Answer:

A 50-mL volumetric cylinder with 0.1-mL accuracy scale should be used for this purpose since three significant figures of accuracy are required.

Explanation:

Hello,

A 50-mL volumetric cylinder with 0.1-mL accuracy scale should be used for this purpose since three significant figures of accuracy are required.

Best regards.

6 0

3 years ago

Read 2 more answers

Enthalpy of <br><br> CH4(g) + 2NO2(g) -> N2(g) + CO2(g) + 2H2O(l)

stira [4]

Answer:

-177.9 kJ.

Explanation:

Use Hess's law. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2Ca(s) + O2(g) → 2CaO(s) ΔH = -1269.8 kJ We need to get rid of the Ca and O2 in the equations, so we need to change the equations so that they're on both sides so they "cancel" out, similar to a system of equations. I changed the second equation. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ The sign changes in the second equation above since the reaction changed direction. Next, we need to multiply the first equation by two in order to get the coefficients of the Ca and O2 to match those in the second equation. We also multiply the enthalpy of the first equation by 2. 2Ca(s) + 2CO2(g) + O2(g) → 2CaCO3(s) ΔH = -1625.6 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ Now we add the two equations. The O2 and 2Ca "cancel" since they're on opposite sides of the arrow. Think of it more mathematically. We add the two enthalpies and get 2CaO(s) + 2CO2(g) → 2CaCO3(s) and ΔH = -355.8 kJ. Finally divide by two to get the given equation: CaO(s) + CO2(g) → CaCO3(s) and ΔH = -177.9 kJ.

5 0

2 years ago