The process of transporting rock material is called

Please help me with this 2 answers i will give brainliest to the more good answer

Fudgin [204]

Answer:

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8 0

2 years ago

Please calculate the mixed volume (Vmix) when 0.300 mol of ethanol and 0.200 mol of water are mixed. Hint: The partial molar vol

Grace [21]

Answer:

If we assume the molar volumes of water and ethanol 17.0 and 57.0 cm³/mol, respectively, Vmix = 20.5 cm³.

Explanation:

The molar volume of a substance is the ratio between the volume and the number of moles of the substance. It represents the volume that 1 mol of it occupies. Because we don't have access to page 24, let's assume the molar volumes of water and ethanol 17.0 and 57.0 cm³/mol, respectively.

The volume of mixture (Vmix) is the sum of the volume of each substance, which is the number of moles multiplied by molar volume, so:

Vmix = 0.300*57 + 0.200*17

Vmix = 17.1 + 3.4

Vmix = 20.5 cm³

7 0

2 years ago

Why can an increase in temperature lead to more effective collisions between reactant particles and an increase in the rate of a

ankoles [38]

An an increase in temperature lead to more effective collisions between reactant particles and an increase in the rate of a chemical reaction because the number of molecules with sufficient energy to react increases. The answer is number 3.

3 0

3 years ago

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What mass of HgO is required to produce 0.692 mol of O2? 2HgO(s) -> 2Hg(l) + O2(g)

Vika [28.1K]

The answer for the following problem is mentioned below.

Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule

Explanation:

Given:

no of moles of the oxygen gas = 0.692

Also given:

2 HgO → 2 Hg + $O_{2}$

where,

HgO represents mercuric oxide

Hg represents mercury

$O_{2}$ represents oxygen

To calculate:

Molar mass of HgO:

Molar mass of HgO = 216 grams

molar mass of mercury (Hg) = 200 grams

molar mass of oxygen (O) =16 grams

HgO = 200 +16 = 216 grams

We know;

2×216 grams of HgO → 1 mole of oxygen molecule

? → 0.692 moles of oxygen molecule

= $\frac{2*216*0.692}{1}$

= 298.944 grams of HgO

Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule

7 0

2 years ago

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If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under

WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

$n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1$

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

$n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol$

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

$PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\ then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L$

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0

2 years ago