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2 years ago

If 2.9g of water is heated from 23.9C to 98.9C, how much heat (in calories) was added to the water?

Chemistry

1 answer:

tensa zangetsu [6.8K]2 years ago

7 0

Answer:

Explanation:

we know that

ΔH=m C ΔT

where ΔH is the change in enthalpy (j)

m is the mass of the given substance which is water in this case

ΔT IS the change in temperature and c is the specific heat constant

we know that given mass=2.9 g

ΔT=T2-T1 =98.9 °C-23.9°C=75°C

specific heat constant for water is 4.18 j/g°C

therefore ΔH=2.9 g*4.18 j/g°C*75°C

ΔH=909.15 j

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Classify the following atom as excited, ground state, or not possible.<br><br> 1s^2 2s^2 2p^6 3p^1

Luda [366]

According to Pauli's exclusion principle, the energy levels of the subshells should be arranged in an increasing manner. The energy of each subshell can be computed. In the sequence, the given subshells are arranged in a correct order. 30p allows 6 electrons at maximum so this configuration is considered as ground state

8 0

3 years ago

A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 degrees celsius,

Airida [17]

Answer:

0.714 liter.

Explanation:

Given:

The balloon initially has a volume of 0.4 liters and a temperature of 20 degrees Celsius.

It is heated to a temperature of 250 degrees Celsius.

Question asked:

What will be the volume of the balloon after he heats it to a temperature of 250 degrees Celsius ?

Solution:

By using:

$PV=nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 0.4 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

0.4 liter = $0.4\times10^{-3}=4\times10^{-4} m^{3}$

And initial temperature of balloon, $T_{1}$ = 20°C = (273 + 20)K

= 293 K

Let the final volume of balloon is $V_{2}$

And a given, final temperature of balloon, $T_{2}$ is 250°C = (273 + 250)K

= 523 K

Now, $V_{1}$ = $KT_{1}$

$4\times10^{-4}=K\times293\ (equation\ 1 )$

$V_{2}$ = $KT_{2}$

$=K\times523\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4\times10^{-4}}{V_{2} } =\frac{K\times293}{K\times523}$

K cancelled by K.

By cross multiplication:

$293V_{2} =4\times10^{-4} \times523\\V_{2} =\frac{ 4\times10^{-4} \times523\\}{293} \\ = \frac{2092\times10^{-4}}{293} \\ =7.14\times10^{-4}m^{3}$

Now convert it into liter with the help of calculation done above.

$7.14\times10^{-4} \times1000\\7.14\times10^{-4} \times10^{3} \\0.714\ liter$

Therefore, the volume of the balloon be after he heats it to a temperature of 250 degrees Celsius is 0.714 liter.

5 0

2 years ago

If a container were to have 24 molecules of C5H12 and 24 molecules of O2 initially, how many total molecules (reactants plus pro

frutty [35]

Answer:

81 molecules

Explanation:

The reaction between C5H12 and O2 is a combustion reaction and is represented by the following equation;

C5H12 + 8O2 --> 5CO2 + 6H2O

The ratio of C5H12 to O2 from the above equation is 1 : 8.

Aplying the conditins of the question; 24 molecules each of C5H12 and O2 we have;

3C5H12 + 24O2 --> 15CO2 + 18H2O

This means we have 24 - 3 = 21 molecules of C5H12 that are unreacted.

Total molecules is given as;

3(C5H12) + 24(O2) + 15(CO2) + 18(H2O) + 21(Unreacted C5H12) = 81 molecules

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2 years ago

Which organelle's function is CORRECTLY described?

laiz [17]

The correct answer is c.

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2 years ago

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Please help me!! (brainliest)

Ymorist [56]

Answer:

Explanation:

6 0

3 years ago