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ale4655 [162]
2 years ago
15

What is the molecular formula of the compound whose molar mass is 90.08 g/mol and empirical formula is CH2O??

Chemistry
1 answer:
KIM [24]2 years ago
7 0
Let (CH2O)n be molecular formula,
(12+1+1+16)n=90.08
30n=90.08
n=3
molecular formula: (CH2O) 3= C3H6O3
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iVinArrow [24]

Answer:

16 mol NaCl.

Explanation:

Do the train track method to cancel out all the units except moles of NaCl on top. Remember one mole of any gas occupies 22.4 L at STP.

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2 years ago
What is the molarity of 2.00 L of a solution that contains 14.6 g NaCl?
White raven [17]

Answer: The molarity of the solution is 0.125 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

moles of NaCl = \frac{\text {given mass}}{\text {Molar mass}}=\frac{14.6g}{58.5g/mol}=0.250mol

Now put all the given values in the formula of molality, we get

Molarity=\frac{0.250mol}{2.00L}=0.125M

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8 0
3 years ago
Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou
Ulleksa [173]

Answer:

4.36~g~XY

Explanation:

In this case, we can start with the reaction:

2X + Y_2~->~2XY

If we check the reaction, we will have 2 X and Y atoms on both sides. So, <u>the reaction is balanced</u>. Now, the problem give to us two amounts of reagents. Therefore, we have to find the <u>limiting reagent</u>. The first step then is to find the moles of each compound using the <u>molar mass</u>:

3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X

4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2

Now, we can <u>divide by the coefficient</u> of each compound (given by the balanced reaction):

\frac{0.04~mol~X}{1}=~0.04

\frac{0.0875~mol~Y_2}{2}=0.04375

The smallest value is for "X", therefore this is our <u>limiting reagent</u>. Now, if we use the <u>molar ratio</u> between "X" and "XY" we can calculate the moles of XY, so:

0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol Y_2 = 48 g Y_2 (therefore 1 mol Y = 24 g Y). With this in mind the <u>molar mass of XY</u> would be 85+24 = 109 g/mol. With this in mind:

0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY

I hope it helps!

6 0
3 years ago
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