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Nastasia [14]
3 years ago
7

How many kJ are 3,340J?

Chemistry
1 answer:
forsale [732]3 years ago
8 0
The answer I believe is 3.340kj.
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Water has a heat of fusion of about 300J/gram and a heat capacity for liquid water of a about 4J/gC. How much heat is needed to
amm1812

Answer:

The total heat required is 3.4 kJ

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

There is a direct proportional relationship between heat and temperature. So, the amount of heat a body receives or transmits is determined by:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case you know;

  • Q= ?
  • c= 4 \frac{J}{g*C}
  • m= 10 g
  • ΔT= Tfinal - Tinitial= 10 C - 0 C= 10 C

Replacing:

Q= 4\frac{J}{g*C} *10 g * 10 C

Solving:

<em>Q1= 400 J</em>

On the other hand, you must determine the heat required to convert 0 ∘ C of ice to 0 ∘ C of liquid water by:

Q2=m*heat of fusion

Q2=10 g* 300 \frac{J}{g}

<em>Q2= 3,000 J</em>

The total heat required is:

Q= Q1 + Q2= 400 J + 3,000 J

Q= 3,400 J= 3.4 kJ (1 kJ= 1,000 J)

<u><em>The total heat required is 3.4 kJ</em></u>

5 0
3 years ago
How many moles of Ammonium Sulfate can be made from 30.0mol of NH3? 2NH3 + H2SO4 --&gt; (NH4)2SO4
BartSMP [9]

Answer: 15.0 moles of (NH_4)_2SO_4 are formed from  30.0 mol of NH_3

Explanation:

The balanced chemical reaction is :

2NH_3+H_2SO_4\rightarrow (NH_4)_2SO_4

According to stoichiometry :

2 moles of NH_3 give = 1 mole of (NH_4)_2SO_4

Thus 30.0 moles of NH_3 will give =\frac{1}{2}\times 30.0=15.0moles  of (NH_4)_2SO_4

Thus 15.0 moles of (NH_4)_2SO_4 are formed from  30.0 mol of NH_3

7 0
3 years ago
How many joules of heat are required to raise the temperature of 174g of gold from 22°C to 85°C? The specific heat of gold is 0.
Slav-nsk [51]
Remembering the equation Q=MCdeltaT where
q=is the amount of heat energy
M=mass
C=specific heat
deltaT= change in temperature

Therefore, using the equation we can substitute values and solve for q.
Q= (15 grams) (0.129 J/(gx°C))(85-22)
Q=(15) ((0.129 J/(gx°C)) (63)
Q=121.9 Joules

The energy needed to raise the temperature of 15 grams of gold from 22 degrees Celsius to 85 degrees Celsius is then 121.9 Joules or 122 Joules (if rounded up).

7 0
2 years ago
After the child exhaled all of the gas, it becomes sick
tensa zangetsu [6.8K]
What is your question about it?
5 0
2 years ago
Read 2 more answers
Suppose that coal of density 1.5 g/cm^3 is pure carbon. (It is, in fact, much more complicated, but this is a reasonable first a
NISA [10]

Answer:

q = -6464.9 kJ

Explanation:

We are given that the heat of combustion is  ∆H° = −394 kJ per mol of carbon.Therefore what we need to do is calculate how many moles of C are in the lump of coal by finding its mass since the density is given.

vol = 5.6 cm x 5.1 cm x 4.6 cm = 131.38 cm³

m = d x v = 1.5 g/cm³ x 131.38 cm³ = 197.06 g

mol C = m/MW = 197.06 g/ 12.01g/mol = 16.41 mol

q =  −394 kJ /mol C x 16.41 mol C = -6464.9 kJ

7 0
3 years ago
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