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3 years ago

Which spheres are part of the earth system

Chemistry

1 answer:

zubka84 [21]3 years ago

8 0

Answer:

Everything in Earth's system can be placed into one of four major subsystems: land, water, living things, or air. These four subsystems are called "spheres." Specifically, they are the "lithosphere" (land), "hydrosphere" (water), "biosphere" (living things), and "atmosphere" (air).

Explanation:

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Compute the freezing point of this solution:

likoan [24]

Answer:

Freezing point = 1.25

Explanation:

If we increase the concentration of the solution, the concentration of H+ does not change.

Convert 2.5% in to decimal

2.5% = 2.5 ÷100

= 0.025

The freezing point = 0.025 × 50

= 1.25

8 0

3 years ago

For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

The reaction

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) Maximun ammount of nitrogen gas => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

8 0

3 years ago

Sixty liters of a gas were collected over water when the barometer read 663 mmhg , and the temperature was 20∘c. what volume wou

lesya [120]

First, let's compute the number of moles in the system assuming ideal gas behavior.

PV = nRT
(663 mmHg)(1atm/760 mmHg)(60 L) = n(0.0821 L-atm/mol-K)(20+273 K)
Solving for n,
n = 2.176 moles

At standard conditions, the standard molar volume is 22.4 L/mol. Thus,

Standard volume = 22.4 L/mol * 2.176 mol = 48.74 L

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3 years ago

An increase in the surface area of reactants in a heterogeneous reaction will result in what?

marusya05 [52]

It will result in an increase in the rate of rxn

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3 years ago

What are the two types of stoichiometry?

julsineya [31]

Two types of stoichiometry are; molar mass and coefficients from balanced equation.

8 0

3 years ago

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