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san4es73 [151]
3 years ago
10

Calculate the amount of heat required to raise the temperature of a 65-g sample of water from 32 ∘C to 65 ∘C. (The specific heat

capacity of water is 4.184 J/(g⋅∘C).
Chemistry
2 answers:
dsp733 years ago
8 0

The amount of heat required is 9.0 kJ.

<em>q = mC</em>Δ<em>T </em>

Δ<em>T</em> = <em>T</em>_f – <em>T</em>_i = 65 °C – 32 °C = 33 °C

<em>q</em> = 65 g × 4.184 J·°C⁻¹g⁻¹ × 33 °C = 9000 J = 9.0 kJ

RUDIKE [14]3 years ago
4 0

Answer:

8974.68 J.

Explanation:

  • The amount of heat required to raise the temperature of water sample can be calculated using the formula:

<em>Q = m.C.ΔT</em>, where,

Q is the amount of heat required to raise the temperature of water sample,

m is the mass of water sample <em>(m = 65.0 g)</em>,

C is the specific heat of water (C = 4.184 J/g.°C),

ΔT is the temperature difference (the final temperature - the initial temperature <em>(ΔT = 65.0 - 32.0 = 33.0 °C)</em>.

∴ The amount of heat required to raise the temperature of water sample Q = m.C.ΔT = (65.0 g) (4.184 J/g.°C) (33.0 °C) = 8974.68 J.

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The electron configuration of a neutral atom is 1s22s22p63s2. Write a complete set of quantum numbers for each of the electrons.
Pavlova-9 [17]

Answer:

The four quantum number for each electron will be:

1s^{2}

n=1;l=0;m=0;s=+\frac{1}{2}/-\frac{1}{2}

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2p^{6}

n=1;l=1;\\m=+1,0.-1 \\s=+\frac{1}{2}/-\frac{1}{2}

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n=3;l=0;m=0;s=+\frac{1}{2}/-\frac{1}{2}

Explanation:

As the element is neutral, the number of protons will be equal to number of electrons which will be the atomic number of the element.

Number of electrons =12

Atomic number = 12

Element : Magnesium

The principal shell is represented by "n"

i) For "s" subshell the value of l =0 (azimuthal quantum number) thus m (magnetic quantum number)= 0

The two electrons in s subshell will have either plus half or minus half spin quantum number

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8 0
3 years ago
At the normal melting point of NaCl, 801 degrees C, its enthalpy of fusion is 28.8 kJ / mol. The density of the solid is 2.165 g
melisa1 [442]

<u>Answer:</u> The increase in pressure is 0.003 atm

<u>Explanation:</u>

To calculate the final pressure, we use the Clausius-Clayperon equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = initial pressure which is the pressure at normal boiling point = 1 atm

P_2 = final pressure = ?

\Delta H = Enthalpy change of the reaction = 28.8 kJ/mol = 28800 J/mol     (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 801^oC=[801+273]K=1074K

T_2 = final temperature = (801+1.00)^oC=802.00=[802+273]K=1075K

Putting values in above equation, we get:

\ln(\frac{P_2}{1})=\frac{28800J/mol}{8.314J/mol.K}[\frac{1}{1074}-\frac{1}{1075}]\\\\\ln P_2=3\times 10^{-3}atm\\\\P_2=e^{3\times 10^{-3}}=1.003atm

Change in pressure = P_2-P_1=1.003-1.00=0.003atm

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Answer:

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