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grigory [225]
4 years ago
10

If 801 J of heat is available, what is the mass in grams of iron (specific heat = 0.45 J/g・°C) that can be heated from 22.5°C to

120.0°C?
Chemistry
1 answer:
inysia [295]4 years ago
3 0

Answer:

The correct answer will be "18.25 g".

Explanation:

The given values are:

Specific heat,

C = 0.45 J/g・°C

Heat involved,

q =  801 J

Temperature,

ΔT = 120.0°C-22.5°C

     = 97.5°C

As we know,

⇒  C = \frac{q}{m \Delta T}

On substituting the given values, we get

⇒  0.45=\frac{801}{m(97.5)}

⇒  m = 18.25 \ g

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Define the term relative atomic mass, A.
Naily [24]
The mass of an atom and the equivalent of to the number of protons and neutrons in the atom
6 0
3 years ago
How much time would it take for 336 mg of copper to be plated at a current of 5.6 A ? Express your answer using two significant
schepotkina [342]

Answer:

1.8 × 10² s

Explanation:

Let's consider the reduction that occurs upon the electroplating of copper.

Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)

We will establish the following relationships:

  • 1 g = 1,000 mg
  • The molar mass of Cu is 63.55 g/mol
  • When 1 mole of Cu is deposited, 2 moles of electrons circulate.
  • The charge of 1 mole of electrons is 96,486 C (Faraday's constant).
  • 1 A = 1 C/s

The time  that it would take for 336 mg of copper to be plated at a current of 5.6 A is:

336mgCu \times \frac{1gCu}{1,000mgCu} \times \frac{1molCu}{63.55gCu} \times \frac{2mole^{-} }{1molCu} \times \frac{94,486C}{1mole^{-}} \times \frac{1s}{5.6C} = 1.8 \times 10^{2} s

3 0
3 years ago
If the percent yield for the following reaction is 75.0%, and 25.0 g of NO₂ are consumed in the reaction, how many grams of nitr
victus00 [196]

Answer:

17.1195 grams of nitric acid are produced.

Explanation:

3NO_2+H_2O\rightarrow 2HNO_3+NO

Moles of nitrogen dioxide :

\frac{25.0 g}{56 g/mol}=0.5434 mol

According to reaction 3 moles of nitrogen dioxides gives 2 moles of nitric acid.

Then 0.5434 moles of nitrogen dioxides will give:

\frac{2}{3}\times 0.5434 mol=0.3623 mol of nitric acid.

Mass of 0.3623 moles of nitric acid :

0.3623 mol\times 63 g/mol=22.8260 g

Theoretical yield = 22.8260 g

Experimental yield = ?

\%Yield=\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100

75\%=\frac{\text{Experimental yield}}{22.8260 g}

Experimental yield of nitric acid = 17.1195 g

7 0
3 years ago
PLEASE HELP I WILL GIVE BRAINLIEST!!!!!
dangina [55]

Answer:

I think the answer is c!

Explanation:

Pure substances are further broken down into elements and compounds. Mixtures are physically combined structures that can be separated into their original components. A chemical substance is composed of one type of atom or molecule. hope this helps you!

8 0
3 years ago
PLEASE HELP ASAP
Lapatulllka [165]

Answer:

See explanation

Explanation:

We must first write the equation of the reaction as follows;

C3H8 + 5O2 ----> 3CO2 + 4H2O

Now;

We obtain the number of moles of C3H8 = 132.33g/44g/mol = 3 moles

So;

1 mole of C3H8 yields 3 moles of CO2

3 moles of C3H8 yields 3 × 3/1 = 9 moles of CO2

We obtain the number of moles of oxygen = 384.00 g/32 g/mol = 12 moles

So;

5 moles of oxygen yields 3 moles of CO2

12 moles of oxygen yields 12 × 3/5 = 7.2 moles of CO2

We can now decide on the limiting reactant to be C3H8

Therefore;

Mass of CO2 produced = 9 moles of CO2 × 44 g/mol = 396 g of CO2

Again;

1 moles of C3H8 yields 4 moles of water

3 moles of C3H8 yields 3 × 4/1 = 12 moles of water

Hence;

Mass of water = 12 moles of water × 18 g/mol = 216 g of water

In order to obtain the percentage yield from the reaction, we have;

b) Actual yield = 269.34 g

Theoretical yield = 396 g

Therefore;

% yield = actual yield/theoretical yield × 100/1

Substituting values

% yield = 269.34 g /396 g × 100

% yield = 68%

8 0
3 years ago
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