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2 years ago

Greenhouse gases in the atmosphere contribute to

Chemistry

1 answer:

lesantik [10]2 years ago

6 0

Answer:

Explanation:

While solar power does produce carbon dioxide emissions in the manufacturing processes required to produce panels and batteries, it does not produce CO2 while converting solar energy to electrical energy.

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A 4.55 l sample of water contains 0.115 g of sodium ions. determine the concentration of sodium ions in ppm if the density of th

Slav-nsk [51]

When one talks about ppm in a liquid solution someone means mg/L so we would not be using the density. This usually means ug/g or mg/kg 0.115 g Na^+ * 10^6 ug/1 g = 115000 ug/g 4.55 L * 1000 mL/1L = 4550 mL Concentration of Na^+ in ppm: 115000 ug/g /4550 mL = 25.27 pm of sodium ion

6 0

3 years ago

Read 2 more answers

For each of the following balanced chemical equations, calculate how many moles and how many grams of each product would be prod

natima [27]

Answer:

(a)

Moles of ammonium chloride = 0.5 mole

Mass of ammonium chloride formed = 26.7455 g

(b)

Mole of $CS_2$ = 0.125 mole

Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g

Mole of $H_2S$ = 0.25 mole

Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g

Explanation:

(a)

For the first reaction:-

$NH_3_{(g)}+HCl_{(g)}\rightarrow NH_4Cl_{(s)}$

The mole ratio of the reactants = 1 : 1

0.5 moles of ammonia react with 0.5 moles of hydrochloric gas to give 0.5 moles of ammonium chloride

So, Moles of ammonium chloride formed = 0.5 moles

Molar mass of ammonium chloride = 53.491 g/mol

Mass = Moles * Molar mass = 0.5 * 53.491 g = 26.7455 g

(b)

For the first reaction:-

$CH_4_{(g)}+4S_{(s)}\rightarrow CS_2_{(l)}+2H_2S_{(g)}$

The mole ratio of the reactants = 1 : 4

It means

0.5 moles of methane react with 2.0 moles of sulfur to give 0.5 moles of Carbon disulfide and 1.0 moles of hydrogen sulfide gas.

But available moles of S = 0.5 moles

Limiting reagent is the one which is present in small amount. Thus, S is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

4 moles of S produces 1 mole of $CS_2$

Thus,

0.5 moles of S produces $\frac{1}{4}\times 0.5$ mole of $CS_2$

Mole of $CS_2$ = 0.125 mole

Molar mass of $CS_2$ = 76.139 g/mol

Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g

4 moles of S produces 2 moles of $H_2S$

Thus,

0.5 moles of S produces $\frac{2}{4}\times 0.5$ mole of $H_2S$

Mole of $H_2S$ = 0.25 mole

Molar mass of $H_2S$ = 34.1 g/mol

Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g

7 0

3 years ago

AlCl3 + Na NaCl + Al Did Na change oxidation number?

Ilya [14]

$Al^{+III}Cl^{-I} _{3}+Na^{0}\rightarrow Na^{+I}Cl^{-I} + Al^{0}\\\\
Yes$

3 0

3 years ago

Explain why the product at the negative electrode is not always a metal.

Elena-2011 [213]

Answer:

Whether you get the metal or hydrogen during electrolysis depends on the position of the metal in the reactivity series: the metal will be produced if it is less reactive than hydrogen. hydrogen will be produced if the metal is more reactive than hydrogen.

3 0

3 years ago

Why is this reaction considered to be exothermic? a Because energy difference A is greater than energy difference C. b Because e

Pani-rosa [81]

Answer:

we would need a bit more info for this

Explanation:

4 0

3 years ago