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Calculate the volume in milliliters of a 1.3M zinc nitrate solution that contains 100.g of zinc nitrate ZnNO32. Be sure your ans

zheka24 [161]

Answer:

The volume is 406 mL

Explanation:

Step 1: Data given

Molarity of a zinc nitrate solution = 1.3 M

Mass of zinc nitrate = 100 grams

Molar mass of Zn(NO3)2 = 189.36 g/mol

Step 2: Calculate moles Zn(NO3)2

Moles Zn(NO3)2 = mass Zn(NO3)2 / molar mass Zn(NO3)2

Moles Zn(NO3)2 = 100.0 grams / 189.36 g/mol

Moles Zn(NO3)2 = 0.5281 moles

Step 3: Calculate volume

Molarity = moles / volume

Volume = moles / molarity

Volume = 0.5281 moles / 1.3 M

Volume = 0.406 L = 406 mL

The volume is 406 mL

3 0

3 years ago

Given a compound with a mw of 169 amu having 13.60% na, 8.29% n, 35.51% c, 4.77% h, and 37.85% o, determine the number of each a

Arisa [49]

1) Percent composition

Na: 13.60%,
N: 8.29%
C: 35.51%
H: 4.77%
O: 37.85%
 -------------
 100.02%

2) Convert mass percent to molar composition

For that, divide the percent of each element by its atomic mass.


Na: 13.60 / 22.99 = 0.5916
N: 8.29 / 14.01 = 0.5917
C: 35.51 / 12.01 = 2.9567
H: 4.77 / 1.01 = 4.7228
O: 37.85 / 16.00 = 2.3656

3) Divide each number by the smallest one:


Na: 0.5916 / 0.5916 = 1.0
N: 0.5917 / 0.5916 = 1.0
C: 2.9567 / 0.5916 = 5.0
H: 4.7228 / 0.5916 = 8.0
O: 2.3656 / 0.5916 = 4.0

4) State the empirical formula: Na N C5 H8 O4

5) Calculate the mass of the empirical formula:

Multiplicate the number of each atoms times the atomic mass corresponding atomic mass of the atom.

1 * 22.99 + 1 * 14.01 + 5 * 12.01 + 8 * 1.01 + 4 * 16.00 = 169.13

6) Divide the molar mass by the mass of the empirical formula:

169 / 169.13 = 1

7) Conclusion: the molecular formula is Na N C5 H8 O4, so the number of atoms present in the compound formula are:

Na: 1
N: 1
C: 5
N: 8
O: 4

5 0

3 years ago

Read 2 more answers

When 0.100 mol of carbon is burned in a closed vessel with8.00

antoniya [11.8K]

Answer : The mass of carbon monoxide form can be 2.8 grams.

Solution : Given,

Moles of C = 0.100 mole

Mass of $O_2$ = 8.00 g

Molar mass of $O_2$ = 32 g/mole

Molar mass of CO = 28 g/mole

First we have to calculate the moles of $O_2$ .

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2C+O_2\rightarrow 2CO$

From the balanced reaction we conclude that

As, 2 mole of $C$ react with 1 mole of $O_2$

So, 0.1 moles of $C$ react with $\frac{0.1}{2}=0.05$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $C$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO$