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3 years ago

18 g of copper is mixed with silver nitrate in water. How much copper ll nitrate will formed?

Chemistry

1 answer:

Lorico [155]3 years ago

7 0

Answer:

Mass = 112.54 g

Explanation:

Given data:

Mass of copper = 18 g

How much copper(II) nitrate formed = ?

Solution:

Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag

Number of moles of copper:

Number of moles = mass/ molar mass

Number of moles = 18 g/ 29 g/mol

Number of moles = 0.6 mol

Now we will compare the moles of Cu with Cu(NO₃)₂ .

Cu : Cu(NO₃)₂

1 : 1

0.6 : 0.6

Mass of Cu(NO₃)₂ :

Mass = number of moles × molar mass

Mass = 0.6 mol × 187.56 g/mol

Mass = 112.54 g

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What is the change in the freezing point of water when 35.0 g of sucrose is

frez [133]

From the calculations, we can see that, the change in the freezing point is -0.634°C.

<h3>What is freezing point?</h3>

The term freezing point refers to the temperature at which a liquid is changed to solid.

Given that;

ΔT = K m i

Number of moles sucrose = 35.0 g/ 342.30 g/mol = 0.1 moles

molality = 0.1 moles/ 300.0 * 10^-3 Kg

= 0.33 m

Thus;

ΔT = -1.86°C/mol * 0.33 m * 1

= -0.634°C

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2 years ago

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3 years ago

The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calculate the volume of CO2 genera

nevsk [136]

Answer:

The volume of carbon dioxide gas generated 468 mL.

Explanation:

The percent by mass of bicarbonate in a certain Alka-Seltzer = 32.5%

Mass of tablet = 3.45 g

Mass of bicarbonate = $3.45 g\times \frac{32.5}{100}=1.121 mol$

Moles of bicarbonate ion = $\frac{1.121 g/mol}{61 g/mol}=0.01840 mol$

$HCO_3^{-}(aq)+HCl(aq)\rightarrow H_2O(l)+CO_2(g)+Cl^-(aq)$

According to reaction, 1 mole of bicarbonate ion gives with 1 mole of carbon dioxide gas , then 0.01840 mole of bicarbonate ion will give:

$\frac{1}{1}\times 0.01840 mol=0.01840 mol$ of carbon dioxide gas

Moles of carbon dioxide gas n = 0.01840 mol

Pressure of the carbon dioxide gas = P = 1.00 atm

Temperature of the carbon dioxide gas = T = 37°C = 37+273 K=310 K

Volume of the carbon dioxide gas = V

$PV=nRT$ (ideal gas equation)

$V=\frac{nRT}{P}=\frac{0.01840 mol\times 0.0821 atm L/mol K\times 310 K}{1.00 atm}=0.468 L$

1 L = 1000 mL

0.468 L =0.468 × 1000 mL = 468 mL

The volume of carbon dioxide gas generated 468 mL.

5 0

3 years ago

What is the mass in grams of 1.00 x 10 24 atoms of Mn?

Alex17521 [72]

<h3>Answer:</h3>

91.2 g Mn

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.00 × 10²⁴ atoms Mn

<u>Step 2: Identify Conversions</u>

Avogadro's Numer

[PT] Molar Mass of Mn - 54.94 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 1.00 \cdot 10^{24} \ atoms \ Mn(\frac{1 \ mol \ Mn}{6.022 \cdot 10^{23} \ atoms \ Mn})(\frac{54.94 \ g \ Mn}{1 \ mol \ Mn})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 91.2321 \ g \ Mn$