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2 years ago

Under which conditions of temperature and pressure would helium behave most like an ideal gas?

2 answers:

8 0

At conditions 50 K and 20 kPa of temperature and pressure would helium behave most like an ideal gas. The answer is number 1. This is because the temperature is so low and the pressure is low too.

xenn [34]2 years ago

5 0

Answer : The correct option is, (3) 750 K and 20 kPa

Explanation :

The conditions for ideal gas are :

Ideal gas are those gas that has no intermolecular attractions.

Ideal gas are those gas that have negligible volume.

The ideal gas equation is,

$PV=nRT$

The conditions for real gas are :

Real gas are those gas that have intermolecular attractions.

Real gas are those gas that have volume.

The real gas equation is,

$(P+\frac{an^2}{V^2})(V-nb)=nRT$

A real gas behave ideally at high temperature and low pressure condition.

From the given options, option (3) have high temperature and low pressure is the correct option.

Hence, at 750 K and 20 kPa conditions of temperature and pressure does a sample of helium behave most like an ideal gas.

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As Danny was pouring cereal for his breakfast, he noticed that the cereal box says that the cereal contains 5 milligrams of iron

Vinil7 [7]

Answer:

Correct, because B it is reported to the nearest miligram

Explanation:

4.6 rounded up is 5

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2 years ago

Driving cars lowers the pH of the oceans by _______.

Anna [14]

Correct answer: a. releasing CO2 that dissolves and forms acid in the oceans

The fuels used in automobiles release gases like carbon dioxide, carbon monoxide, oxides of nitrogen and sulfur. Carbon dioxide when dissolved in water forms carbonic acid. So, when the usage of cars is high, these emissions of carbon-dioxide into the atmosphere increase and this leads to the lowering of pH of the oceans as the carbon dioxide present in higher amounts in to atmosphere diffuses into the oceanic waters and form carbonic acid which makes the ocean slightly acidic.

$CO_{2}(g) + H_{2}O (l) H_{2}CO_{3}(aq)$

6 0

2 years ago

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The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t

Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

$2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?$

We have :

Enthalpy changes of formation of following s:

$\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol$

$\Delta H_{f,CuO}=-157.3 kJ/mol$

$\Delta H_{f,NO_2}= 33.2 kJ/mol$

$\Delta H_{f,O_2}= 0 kJ/mol$ (standard state)

$\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]$

The equation for the enthalpy change of the given reaction is:

$\Delta H_{rxn}$ =

$=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})$

$\Delta H_{rxn}$ =

$(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)$

$\Delta H_{rxn}=424 kJ$

The enthalpy change for the given reaction is 424 kJ.

6 0

2 years ago

a block of iron has a mass of 826g. what is the mass of a block of magnesium that has the same volume as the block of iron? the

fiasKO [112]

Answer is: the mass of a block of magnesium is 177.75 grams.

m(Fe) = 826 g.

d(Fe) = 7.9 g/cm³.

1) Calculate volume of iron and magnesium:

d(Fe) = m(Fe) ÷ V(Fe).

V(Fe) = m(Fe) ÷ d(Fe).

V(Fe) = 826 g ÷ 7.9 g/cm³.

V(Fe) = V(Mg) = 104.56 cm³.

2) Calculate mass of magnesium:

m(Mg) = V(Mg) · d(Mg).

m(Mg) = 104.56 g/cm³ · 1.7 g/cm³.

m(Mg) = 177.75 g.

8 0

2 years ago

For no2−, write an equation that shows how the anion acts as a base. express your answer as a chemical equation. identify all of

katrin [286]

Hello!

Bases are defined by Arrhenius as substances which release OH⁻ ions when dissolved in water. NO₂⁻ complies with this definition by the chemical reaction that is shown below:

NO₂⁻(aq) + H₂O (l) HNO₂ (aq) + OH⁻(aq)

Have a nice day!

5 0

2 years ago

Read 2 more answers