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erma4kov [3.2K]
4 years ago
8

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

wer has the correct number of significant digits.
Chemistry
1 answer:
zheka24 [161]4 years ago
3 0

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

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Answer:

Single Displacement reaction  

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2KI + Cl2 → 2KCl + I2

CuSO4 + Zn → ZnSO4 + Cu

Double displacement reaction

In a double displacement reaction, two atoms or a group of atoms switch places to form new compounds.

Precipitate is formed.

Salt solutions of two different metals react with each other.

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Na2SO4 + BaCl2 → BaSO4 + 2NaCl

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What is the difference between metallic and non metallic oxides?​
Sveta_85 [38]

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Explanation:

have a great day ahead

tC

4 0
3 years ago
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Fumaric acid, which occurs in many plants, contains, by mass, 41.4% carbon, 3.47% hydrogen, and 55.1% oxygen. The molecular mass
lukranit [14]

Answer:

Explanation:

C = 41.4/12 = 3.43

H = 3.47/1 = 3.47

O = 55.1/16 =3.44

CHO is the skeletal formula (divide each by the lowest number above). The results are close enough to 1 to be 1.

(CHO)_x = 116

C + H + O = 29

(29) _ x = 116

x = 116/29

x = 4

So there area 4 carbons 4 hydrogens and 4 oxygens.

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3 0
3 years ago
A substance X contains 10 gram of calcium carbonate calculate the number of mole of calcium carbonate present in X ​
Tasya [4]

\LARGE{ \boxed{  \rm{ \red{Required \: answer}}}}

☃️ Chemical formulae ➝ \sf{CaCO_3}

<h3><u>How to find?</u></h3>

For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.

\boxed{ \sf{No. \: of \: moles =  \frac{given \: weight}{molecular \: weight} }}

<h3><u>Solution:</u></h3>

Atomic weight of elements:

Ca = 40

C = 12

O = 16

❍ Molecular weight of \sf{CaCO_3}

= 40 + 12 + 3 × 16

= 52 + 48

= 100 g/mol

❍ Given weight: 10 g

Then, no. of moles,

⇛ No. of moles = 10 g / 100 g mol‐¹

⇛ No. of moles = 0.1 moles

☄ No. of moles of Calcium carbonate in that substance = <u>0.1 moles</u>

<u>━━━━━━━━━━━━━━━━━━━━</u>

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4 years ago
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