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Pani-rosa [81]
3 years ago
10

What is the Molarity of a solution with 1.8 mil KCl dissolved to a volume of 1.5L

Chemistry
1 answer:
vova2212 [387]3 years ago
8 0

Answer:1.2M

Explanation:

Molarity=number of moles ➗ volume in liters

Molarity=1.8 ➗ 1.5

Molarity=1.2M

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Compute the following and give your answer on scientific notation
Nata [24]

Answer:

6*10^-3

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Procedure in the attached file

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A hypothesis explains what the scientist thinks will happen during the experiment.
igor_vitrenko [27]

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explained: I took the test

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Which of the elements shown will not form ions, and why will they not do so?
grigory [225]

No elements visible!

Ions form between metals and non-metals.

Hope this helps!

6 0
2 years ago
Iron has a density of 7.87 g/cm3. What is the volume in cm3 of 3.729 g of iron?
KIM [24]

If iron has a density of 7.87g/cm³ and a mass of 3.729g, then the volume of iron is 0.474cm³

HOW TO CALCULATE VOLUME:

  • The volume of a substance can be calculated by dividing the mass by its density. That is;

Volume (mL) = mass (g) ÷ density (g/mL)

  • The density of iron is given as 7.87g/cm³ while its mass is 3.729g of iron. Hence, the volume can be calculated as follows:

Volume = 3.729 ÷ 7.87

Volume = 0.474cm³

Therefore, the volume of iron is 0.474cm³

Learn more: brainly.com/question/2040396?referrer=searchResults

6 0
2 years ago
112 g of aluminum carbide react with 174 g water to produce methane and aluminum hydroxide in the reaction shown below.
dolphi86 [110]

<u>Answer:</u> 4.999 moles of excess reactant will be left over.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

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

Given mass of aluminium carbide = 112 g

Molar mass of aluminium carbide = 143.96 g/mol

Putting values in equation 1:

\text{Moles of aluminium carbide}=\frac{112g}{143.96g/mol}=0.778mol

For the given chemical reaction:

2Al_4C_3(s)+12H_2O(l)\rightarrow 3CH_4(g)+4Al(OH)_3(s)

By the stoichiometry of the reaction:

2 moles of aluminium carbide reacts with 12 moles of water

So, 0.778 moles of aluminium carbide will react with = \frac{12}{2}\times 0.778=4.668 mol of water

Given mass of water = 174 g

Molar mass of water = 18 g/mol

Putting values in equation 1:

\text{Moles of water}=\frac{174g}{18g/mol}=9.667mol

Moles of excess reactant (water) left = 9.667 - 4.668 = 4.999 moles

Hence, 4.999 moles of excess reactant will be left over.

8 0
2 years ago
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