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2 years ago

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3.00 moles of NO2 have a mass of

1 answer:

ExtremeBDS [4]2 years ago

6 0

When solving for the mass of a compound when you’re given the number of moles present, you need to know the molar mass (how many grams there are in a mole of that compound).

In this case, we can make the following equation:

3.00(14.01 + 2(16.00))

In Nitrogen, the molar mass is 14.01 grams per mole, and Oxygen is 16.01 grams per mole.

However, because there are 2 oxygen atoms present per molecule, you must multiply it by two in order to solve for the molar mass.

Anyway:
3.00(14.01 + 2(16.00))

Simplify:
3.00(14.01 + 32.00)
3.00(46.01)

Multiply:
3.00(46.01) = 138.03

3.00 moles of NO2 has a mass of 138.03 grams.
-T.B.

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Starting with 1.5052g of BaCl2•2H2O and excessH2SO4, how many grams of BaSO4 can be formed?

muminat

1.5052g BaCl2.2H2O => 1.5052g / 274.25 g/mol = 0.0054884 mol
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6 0

2 years ago

Read 2 more answers

A sample of bismuth weighing 0.687 g was converted to bismuth chloride by reacting it first with HNO3, and then with HCI, follow

alisha [4.7K]

Answer:

The empirical formula is BiCl3

% Bi = 66.27 %

Explanation:

Step 1: Data given

Mass of bismuth = 0.687 grams

Mass of bismuth chloride produced = 1.032 grams

Molar mass of bismuth = 208.98 g/mol

Molar mass of bismuth chloride = 315.33 g/mol

Step 2: The balanced equation

Step 3: Calculate moles of Bi

Moles Bi = mass Bi / molar mass Bi

Moles Bi = 0.687 grams / 208.98 g/mol

Moles Bi = 0.00329 moles

Step 4: Calculate moles of Cl

Mass of Cl = 1.032 - 0.687 = 0.345 moles

Moles Cl = 0.345 moles / 35.45 g/mol

Moles Cl = 0.00973 moles Cl

Step 5: Calculate mol ratio

We divide by the smaller number of moles:

Bi: 0.00329 / 0.00329 = 1

Cl: 0.0097The empirical formula is BiCl33/0.00329 = 3

Step 6: Calculate molar mass of BiCl3

Molar mass = 208.98 + 3*35.45 = 315.33 g/mol

Step 7: Calculate percent of Bi

% Bi = (208.98 / 315.33) * 100%

% Bi = 66.27 %

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2 years ago

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Explanation:

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2 years ago

If the concentration of a reactant is tripled (all other things remain constant), and the reaction rate increases nine times, wh

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2 years ago

A voltaic cell is constructed with two silver-silver chloride electrodes, where the half-reaction is AgCl (s) + e− → Ag (s) + Cl

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The half-cell reaction is:

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In this case, the cathode and anode both are same. So, $E^o_{cell}$ is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}$

where,

n = number of electrons in oxidation-reduction reaction = 1

$E_{cell}$ = ?

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$[Cl^{-}{concentrated}]$ = 2.22 M

Now put all the given values in the above equation, we get:

$E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}$

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Therefore, the cell emf for this cell is 0.118 V

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2 years ago