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2 years ago

Element Molar Mass (g/mol)

Chemistry

2 answers:

Scorpion4ik [409]2 years ago

7 0

The number of moles of ethanol the chemist will use is 0.65 moles.

A mole is the atomic weight of a molecule of chemical in grams

To calculate the number of moles of ethanol the chemist will use, we apply the formula below.

Formula:

n = R'/M.............. Equation 1

Where:

n = number of moles of ethanol
m = mass of ethanol
M = molar mass of ethanol

From the question,

Given:

R' = 30 g
M = [(12.01)+(1.01×3)+(12.01)+(1.01×2)+(16.0)+(1.01)] = 46.08 g/mol

Substitute these values into equation 1

n = 30/46.08
n = 0.65 moles.

Hence, The number of moles of ethanol the chemist will use is 0.65 moles.

Learn more about moles here: brainly.com/question/15356425

artcher [175]2 years ago

6 0

The number of moles of ethanol the chemist will use in the experiment involving 30g of ethanol is 0.65moles.

<h3>How to calculate number of moles?</h3>

The number of moles of a substance can be calculated by dividing the mass of the substance by its molar mass. That is;

no. of moles = mass ÷ molar mass

According to this question, a chemist will use a sample of 30 g of ethanol (CH3CH2OH) in an experiment. The number of moles can be calculated as follows:

Molar mass of ethanol = 12(2) + 1(5) + 17 = 46g/mol

no of moles = 30g ÷ 46g/mol

no. of moles = 0.65moles

Therefore, the number of moles of ethanol the chemist will use in the experiment involving 30g of ethanol is 0.65moles.

Learn more about moles at: brainly.com/question/1458253

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Explanation:

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$H_2CO_3(aq)\rightarrow 2H^+(aq)+CO_3^{2-}(aq)$ ..[1]

In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:

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From one:[1] ,[2] and [3]:

$2H^+(aq)+CO_3^{2-}(aq)+2K^+(aq)+2OH^{-}(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)+H_2O(l)$

Cancelling common ions on both sides to get net ionic equation :

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2CO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

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Putting values in above equation, we get:

$M_1=\frac{1\times 3.840 M\times 20.0 mL}{2\times 50.2 mL}=0.765 M$

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