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ivanzaharov [21]
3 years ago
12

Two easy uses of mixture​

Chemistry
2 answers:
padilas [110]3 years ago
5 0

Answer:

Explanation:

Here are a few more examples:

1. Smoke and fog (Smog)

2. Dirt and water (Mud)

lyudmila [28]3 years ago
4 0

Explanation:

it helps to make juices.

It helps to make concentrated acid into dilute acid.

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A single covalent bond is made up of​
SIZIF [17.4K]

Answer:

In chemistry, a single bond is a chemical bond between two atoms involving two valence electrons. That is, the atoms share one pair of electrons where the bond forms. Therefore, a single bond is a type of covalent bond.

Explanation:

(copied from Google)

4 0
3 years ago
Mechanism with a small activation energy or one with large activation energy​
tangare [24]

Answer:

Rate depends on the rate constant. The rate constant depends on temperature and activation energy. If you have lower activation energy the rate will be higher. This is why catalysts are added since catalysts provide an alternate pathway that requires lower activation energy and catalysts are added to increase the rate of reaction.

Explanation:

This is only the answer if you were asking:

"Which corresponds to the faster rate: a mechanism with a small activation energy or one with a large activation energy?"

Thats what I understood about your question.

8 0
3 years ago
Chemical formula of binary compound a of oxygen acidic<br>​
Reika [66]

A binary compound of oxygen with another element is called oxide. An oxide is a binary compound of oxygen and another element. Oxygen combines with metals and non-metals to form respective oxides.

8 0
3 years ago
The solubility of calcium sulfate at 30°c is 0.209 g/100 ml solution. calculate its ksp.
inessss [21]
Answer is: Ksp for calcium sulfate is 2.36·10⁻⁴.
Balanced chemical reaction (dissociation):
CaSO₄(s) → Ba²⁺(aq) + SO₄²⁻(aq).
m(CaSO₄) = 0.209 g.
n(CaSO₄) = m(CaSO₄) ÷ M(CaSO₄).
n(CaSO₄) = 0.209 g ÷ 136.14 g/mol.
n(CaSO₄) = 0.00153 mol.
s(CaSO₄) = n(CaSO₄) ÷ V(CaSO₄).
s(CaSO₄) = 0.00153 mol ÷ 0.1 L = 0.0153 M.
Ksp = [Ca²⁺] · [SO₄²⁻].
[Ca²⁺] = [SO₄²⁻] = s(CaSO₄).
Ksp = (0.0153 M)² = 2.36·10⁻⁴.
8 0
2 years ago
Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. If 2.59 g of water is produc
max2010maxim [7]

<u>Answer:</u> The percent yield of the water is 31.98 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For methane:</u>

Given mass of methane = 6.58 g

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of methane}=\frac{6.58g}{16g/mol}=0.411mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 14.4 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{14.4g}{32g/mol}=0.45mol

The chemical equation for the combustion of methane is:

CH_4+2O_2\rightarrow CO_2+2H_2O

By Stoichiometry of the reaction:

2 moles of oxygen gas reacts with 1 mole of methane

So, 0.45 moles of oxygen gas will react with = \frac{1}{2}\times 0.45=0.225mol of methane

As, given amount of methane is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction

2 moles of oxygen gas produces 2 moles of water

So, 0.45 moles of oxygen gas will produce = \frac{2}{2}\times 0.45=0.45 moles of water

  • Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 0.45 moles

Putting values in equation 1, we get:

0.45mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.45mol\times 18g/mol)=8.1g

  • To calculate the percentage yield of water, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of water = 2.59 g

Theoretical yield of water = 8.1 g

Putting values in above equation, we get:

\%\text{ yield of water}=\frac{2.59g}{8.1g}\times 100\\\\\% \text{yield of water}=31.98\%

Hence, the percent yield of the water is 31.98 %

4 0
3 years ago
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