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3 years ago

How many moles are in 20.0 g of water

1 answer:

7 0

Mass = mr x moles

Therefore we rearrange to get

Moles = mass/ mr

Water is H2O so the molecular mass values are
H: 1 , O: 16 (from the periodic table).

We have 2 Hs and 1 O so 1 + 1 + 16 = 18

Mass/ mr = 20/18 = 1.11 moles of water

Hope this helps, feel free to ask if you have any questions :)

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Choose the options below that are true.

Assoli18 [71]

Answer:

The options (A) -The rate law for a given reaction can be determined from a knowledge of the rate-determining step in that reaction's mechanism. and (C) -The rate laws of bimolecular elementary reactions are second order overall ,is true.

Explanation:

(A) -The rate law can only be calculated from the reaction's slowest or rate-determining phase, according to the first sentence.

(B) -The second statement is not entirely right, since we cannot evaluate an accurate rate law by simply looking at the net equation. It must be decided by experimentation.

(D)-The fourth argument is incorrect. We must track the rates of and elementary phase that is following the reaction in order to determine the rate.

Therefore , the first and third statement is true.

5 0

3 years ago

In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very

SVETLANKA909090 [29]

Answer: The final concentration of potassium nitrate is $5.70\times 10^{-6}M$

Explanation:

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

$\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$ .......(1)

Calculating for first dilution:

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL$

Putting values in equation 1, we get:

$7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M$

Calculating for second dilution:

$M_2\text{ and }V_2$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_3\text{ and }V_3$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL$

Putting values in equation 1, we get:

$1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M$

Hence, the final concentration of potassium nitrate is $5.70\times 10^{-6}M$

8 0

3 years ago

Raw materials used in the manufacturing of indigenous soap

Harrizon [31]

Answer:

Fat

Alkali

Explanation:

Fat and alkali are the two primary raw materials needed to manufacture soap.

Sodium hydroxide or potassium hydroxide is generally used as an alkali. The use of alkali depends on the intended application of the soap.

Raw animal fat was used in the past but these days, processed fat is used in the soap manufacturing process. Vegetable fats ( e.g, palm oil, olive oil, coconut oil) are also being used in soap manufacturing.

Additives are also used to enrich the color and texture of the soap.

5 0

3 years ago

Lab Report: Paper Chromatography

vampirchik [111]

Answer:

the answer is "I HOPE IT HELPS"

7 0

3 years ago

State the definition of the partial molar Gibbs energy.

balu736 [363]

Explanation :

As we know that the Gibbs free energy is not only function of temperature and pressure but also amount of each substance in the system.

$G=G(T,P,n_1,n_2)$

where,

$n_1\text{ and }n_2$ is the amount of component 1 and 2 in the system.

Partial molar Gibbs free energy : The partial derivative of Gibbs free energy with respect to amount of component (i) of a mixture when other variable $(T,P,n_j)$ are kept constant are known as partial molar Gibbs free energy of $i^{th}$ component.