All categories

4 years ago

Determine the heat needed to warm 25.3 g of copper from 22 degrees celsius to 39 degrees celsius.

Chemistry

1 answer:

Serggg [28]4 years ago

5 0

Answer:

The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.

Explanation:

$Q=mc\Delta T$

Where:

Q = heat absorbed or heat lost

c = specific heat of substance

m = Mass of the substance

ΔT = change in temperature of the substance

We have mass of copper = m = 25.3 g

Specific heat of copper = c = 0.385 J/g°C

ΔT = 39°C - 22°C = 17°C

Heat absorbed by the copper :

$Q=25.3 g\times 0.385 J/g^oC\times 17^oC=165.59 J$

The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.

You might be interested in

When chlorobenzene reacts with Mg in ether followed by CO2 and neutralization with dilute HCl, __________ will be formed.

allsm [11]

Answer:

c. benzoic acid

Explanation:

The given reaction is an example of a Grignard reaction:

When chlorobenzene (C₆H₅Cl) reacts with Mg in ether, an intermediate is formed (C₆H₅MgCl).

Said intermediate then reacts with CO₂ producing a benzoic acid salt (C₆H₅CO₂X), this salt is then neutralized with dilute HCl producing benzoic acid (C₆H₅CO₂H).

4 0

3 years ago

Assume that concentrated aqueous NH3 has a density of 0.252 g/mL (0.252 g of NH3 per mL of liquid). Calculate the volume of NH3

Kobotan [32]

The question is incomplete, here is the complete question:

Assume that concentrated aqueous NH₃ has a density of 0.252 g/mL (0.252 g of NH₃ per mL of liquid). Calculate the volume of NH₃ required to contain 0.442 mol?

<u>Answer:</u> The volume of ammonia required is 29.82 mL

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of ammonia = 17 g/mol

Moles of ammonia = 0.442 moles

Putting values in above equation, we get:

$0.442mol=\frac{\text{Mass of ammonia}}{17g/mol}\\\\\text{Mass of ammonia}=(0.442mol\times 17g/mol)=7.514g$

To calculate the volume of ammonia, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of ammonia = 0.252 g/mL

Mass of ammonia = 7.514 g

Putting values in above equation, we get:

$0.252g/mL=\frac{7.514g}{\text{Volume of ammonia}}\\\\\text{Volume of ammonia}=\frac{7.514g}{0.252g/mL}=29.82mL$

Hence, the volume of ammonia required is 29.82 mL

4 0

3 years ago

If you use one more of N2 how many moles of NH3 could be produced?

mote1985 [20]

Hey there!

2 moles will be produced.

In N₂ there are 2 nitrogen atoms. In NH₃ there is 1 nitrogen atom.

So, there will be twice as many moles produced because there will be twice as many molecules.

Hope this helps!

8 0

4 years ago

Apply scientific knowledge and understanding to determine how many grams of carbon dioxide are produced from the combustion of 2

lidiya [134]

Answer:

749 grams CO₂

Explanation:

To find the amount of carbon dioxide produced, you need to (1) convert grams C₃H₈ to moles C₃H₈ (via molar mass from periodic table), then (2) convert moles C₃H₈ to moles CO₂ (via mole-to-mole ratio via reaction coefficients), and then (3) convert moles CO₂ to grams CO₂ (via molar mass from periodic table). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units. The desired unit should be in the numerator. The final answer should have 3 significant figures because the given value (250. grams) has 3 sig figs.

Molar Mass (C₃H₈): 3(12.01 g/mol) + 8(1.008 g/mol)

Molar Mass (C₃H₈): 44.094 g/mol

1 C₃H₈ + 5 O₂ ---> 3 CO₂ + 4 H₂O

Molar Mass (CO₂): 12.01 g/mol + 2(16.00 g/mol)

Molar Mass (CO₂): 44.01 g/mol

250. g C₃H₈ 1 mole C₃H₈ 3 moles CO₂ 44.01 g
------------------ x ---------------------- x ---------------------- x -------------------- =
44.094 g 1 mole C₃H₈ 1 mole CO₂

= 749 grams CO₂

6 0

2 years ago

Oxygen gas is collected at a pressure of 1.21 atm in a container which has a volume of 10.0 L. What temperature (in Kelvin) must

stiv31 [10]

Answer : The temperature of gas is, 295 K

Explanation :

To calculate the temperature of gas we are using ideal gas equation:

$PV=nRT$