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PIT_PIT [208]
2 years ago
5

A chemist determined by measurements that 0.050 moles of tin participated in a chemical reaction. Calculate the mass of tin that

participated in the chemical reaction.
Be sure your answer has the correct number of significant digits.
Chemistry
2 answers:
stepladder [879]2 years ago
8 0

Answer:

Explanation:

MM of I2 = 2 (127 g) = 254 g/mol

0.065 mol I2 x 254g I₂/ 1 mol I₂  = 16.5 g I2

TEA [102]2 years ago
5 0
Yes it’s the right Answer!!!
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GaryK [48]
The term which is used is homogeneous.
when sugar is completely dissolved in the water, the mixture or solution homogeneous, both in same phase and same uniform texture that is liquid.
There two types of mixtures are heterogeneous and homogeneous in different phases.
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3 0
3 years ago
In a titration, 4.7 g of an acid (HX) requires 32.6 mL of 0.54 M NaOH(aq) for complete reaction. What is the molar mass of the a
katrin2010 [14]

Answer : The molar mass of an acid is 266.985 g/mole

Explanation : Given,

Mass of an acid (HX) = 4.7 g

Volume of NaOH = 32.6 ml = 0.0326 L

Molarity of NaOH = 0.54 M = 0.54 mole/L

First we have to calculate the moles of NaOH.

\text{Moles of }NaOH=\text{Molarity of }NaOH\times \text{Volume of solution}=0.54mole/L\times 0.0326L=0.017604mole

Now we have to calculate the moles of an acid.

In the titration, the moles of an acid will be equal to the moles of NaOH.

Moles of an acid = Moles of NaOH = 0.017604 mole

Now we have to calculate the molar mass of and acid.

\text{Moles of an acid}=\frac{\text{Mass of an acid}}{\text{Molar mass of an acid}}

Now put all the given values in this formula, we get:

0.017604mole=\frac{4.7g}{\text{Molar mass of an acid}}

\text{Molar mass of an acid}=266.985g/mole

Therefore, the molar mass of an acid is 266.985 g/mole

3 0
3 years ago
What mass of copper is deposited when a current of 10.0a is passed through a solution of copper(ii) nitrate for 30.6 seconds?
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Data Given:

Time = t = 30.6 s

Current = I = 10 A

Faradays Constant = F = 96500

Chemical equivalent = e = 63.54/2 = 31.77 g

Amount Deposited = W = ?

Solution:
             According to Faraday's Law,

                                          W  =  I t e / F

Putting Values,
 
                            W  =  (10 A × 30.6 s × 31.77 g) ÷ 96500

                            W  =  0.100 g

Result:
           0.100 g of Cu
²⁺ is deposited.
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