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2 years ago

Which describes a homogeneous mixture?

Chemistry

1 answer:

Lady_Fox [76]2 years ago

6 0

Answer:

A homogeneous mixture is a solid, liquid or gaseous mixture that has the same proportions of its components throughout any given sample. ... An example of a homogeneous mixture is air. In physical chemistry and materials science this refers to substances and mixtures which are in a single phase.

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Hey there! so I have this question in chemistry where they have shown a reaction with Ammonia acid and HCL, so the question to e

lozanna [386]

What the I haven't even learned this yet

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3 years ago

Read 2 more answers

Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

BARSIC [14]

Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of $CH_4$ = 1.79 mole

Moles of $CO_2$ = 1.20 mole

Moles of $He$ = 3.71 mole

Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0

3 years ago

PLEASE HELP ME!!

ivann1987 [24]

Answer: K and Mg

Explanation:

The first one refers to the atomic radius and increases going down and to the left on the periodic table. K is in between Rb and Na.

6 0

2 years ago

Formulate the sequence of steps needed to prepare 250.0 mL of a 0.005 M aqueous solution from solid KMnO4 and pure water.

S_A_V [24]

Answer:

Molarity is a unit that measures how much moles of solute dissolved in a liter of solvent. Molarity expressed using capital M while molarity, a different unit, expressed using lower case m.

We want to make 0.005 M solution which means we need 0.005 moles of KmnO4 per liter of water. First, we have to calculate how many grams of KMnO4 we need for the solution.

We want to make 250ml solution, so the number of moles of KMnO4 we need will be: 0.005 mol/liter *(250 ml * 1liter/1000ml)= 0.005 mol/liter * 1/4 liter = 0.00125 moles

The molecular mass of KMnO4 is 158g/mol, so the mass of KMnO4 we need will be: 0.00125 moles * 158g/mol= 0.1975 grams

We know that we need 0.1975 g of KMnO4, now we weigh them and put it inside a dish. After that, we prepare Erlenmeyer or a volumetric flask filled with water half of the volume needed(125ml). Pour the weighted solute into the flask, stir until all solute dissolved.

Then we add water to the container slowly until its volume reaches the 250ml mark.

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2 years ago

What is the number of protons and electrons in a neutral atom of<br> Lithium?

nadya68 [22]

If you look at the periodic table you will see the atomic number (the smaller number) which is 3. that identifies how many protons there are. since the lithium is neutral that means there’s a same amount of electrons.

therefore there’s 3 electrons and 3 protons

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2 years ago