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2 years ago

Based on the following chemical equation how many hydrogen atoms are present in the products side?

Chemistry

1 answer:

e-lub [12.9K]2 years ago

8 0

Answer:

the answer is 6

Explanation:

there is 3 hydrogen molecules in NH3 and there's 2 molecules of NH3 so in total, there are 6 hydrogen molecules on the products side.

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A mixture of coarse sand and sugar is 45.0 percent sand by mass. 120.0 grams (g) of the mixture is placed in a fine-mesh cloth b

8090 [49]

Answer:

Option E

Explanation:

From the question we are told that:

Amount of sand in percentage $s_p=45%$

Sample size $n=120g$

Note:After being dumped in the river repeatedly the sugar melts away leaving behind the insoluble sand

Generally the equation for Amount of sand content is mathematically given by

$X=n*s_p$

$X=120*\frac{45}{100}$

$X=54g$

Therefore

After drying, the mass of the contents of the bag equals

$X=54g$

Option E

8 0

3 years ago

The noble gases are grouped together in the periodic table. where are the noble gases in the table?

Rasek [7]

For the noble gasses, you get D, but metals are basically everywhere in the periodic table...

6 0

3 years ago

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A sample of carbon -12 has a mass of 6.00 g. How many atoms of carbon -12 are in the sample ?

nevsk [136]

Answer:

3.02e23

Explanation:

carbon molecular weight = 12 g /mol

1 mol = 6.02e23 atoms

your sample, 6 g / 12 g/mol = 0.5 mol

0.5 x 6.02e23 =

3 0

3 years ago

The acceleration of the car with the data in the table above would bem/s2. If the applied force were cut in half, what do you pr

steposvetlana [31]

Answer:

1. 0.33. 2. 0.17

Explanation:

just did it

4 0

3 years ago

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Write a balanced equation for the combustion of C7H16(l) (heptane) -- i.e. its reaction with O2(g) forming the products CO2(g) a

JulsSmile [24]

Answer:

The standard enthalpy of reaction = -4854.7kJ

The difference: ΔH-ΔE = Δ(PV) = Δn.R.T = 9910.288 J ≈ 9.91 kJ

Explanation:

The balanced chemical equation for the combustion of heptane:

C₇H₁₆ (l) + 11 O₂ (g) → 7 CO₂ (g) + 8 H₂O (l)

Given: The standard enthalpy of formation ( $\Delta H _{f}^{\circ }$ ) for: C₇H₁₆ (l) = -187.8 kJ/mol, O₂ (g) = 0 kJ/mol, CO₂ (g) = -393.5 kJ/mol, H₂O (l) = -286 kJ/mol

To calculate the standard enthalpy of reaction ( $\Delta H _{r}^{\circ }$ ) can be calculated by the Hess's law:

$\Delta H _{r}^{\circ } = \left [\sum \nu \cdot\Delta H _{f}^{\circ }(products) \right ] - \left [\sum \nu\cdot\Delta H _{f}^{\circ }(reactants) \right ]$

Here, $\nu$ is the stoichiometric coefficient

⇒ $\Delta H _{r}^{\circ } =$

$\left [ 7\times \Delta H _{f}^{\circ }\left (CO_{2}\right )+ 8\times \Delta H _{f}^{\circ }\left (H_{2}O \right )\right ]$

$- \left [1\times \Delta H _{f}^{\circ }\left (C_{7}H_{16}\right ) +11\times \Delta H _{f}^{\circ }\left (O_{2} \right ) \right ]$

$=\left [ 7\times \left (-393.5 kJ/mol \right )+ 8\times \left (-286 kJ/mol \right )\right ]$

$-\left [1\times \left (-187.8 kJ/mol \right ) +11\times \left (0 kJ/mol \right ) \right ]$

⇒ $\Delta H _{r}^{\circ } = \left [ \left (-2754.5 \right )+ \left (-2288 \right )\right ]\left -[ \left (-187.8 \right ) +\left (0 \right )\right ]$

⇒ $\Delta H _{r}^{\circ } = \left [ -5042.5 ]\left -[ -187.8] = \left ( -4854.7kJ \right )$

To calculate the difference: ΔH-ΔE=Δ(PV)

We use the ideal gas equation: P.V = n.R.T

⇒ ΔH-ΔE=Δ(PV) = Δn.R.T

Given: Temperature:T = 298K, R = 8.314 J⋅K⁻¹⋅mol⁻¹

Δn = number of moles of gaseous products - number of moles of gaseous reactants = (7)- (11) = (-4)

⇒ ΔH-ΔE=Δ(PV) = Δn.R.T = (-4 mol) × (8.314 J⋅K⁻¹⋅mol⁻¹) × (298K) = 9910.288 J = 9.91 kJ (∵ 1 kJ = 1000J )

8 0

3 years ago