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mixer [17]
1 year ago
14

How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C

Chemistry
1 answer:
NemiM [27]1 year ago
3 0

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

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Harvey kept a balloon with a volume of 348 milliliters at 25.0˚C inside a freezer for a night. When he took it out, its new volu
katrin2010 [14]

Answer:

T2=276K

Explanation:

Given:

Initial volume of the balloon V1 = 348 mL

Initial temperature of the balloon T1 = 255C

Final volume of the balloon V2 = 322 mL

Final temperature of the balloon T2 =

To calculate T1 in kelvin

T1= 25+273=298K

Based on Charles law, which states that the volume of a given mass of a ideal gas is directly proportional to the temperature provided that the pressure is constant. It can be applied using the below formula

(V1/T1)=(V2/T2)

T2=( V2*T1)/V1

T2=(322*298)/348

T2=276K

Hence, the temperature of the freezer is 276 K

8 0
2 years ago
Be sure to answer all parts. Draw the structure of a compound of molecular formula C4H8O that has a signal in its 13C NMR spectr
GenaCL600 [577]

Answer:

The possible structures are ketone and aldehyde.

Explanation:

Number of double bonds of the given compound is calculated using the below formula.

N_{db}=N_{c}+1-\frac{N_{H}+N_{Br}-N_{N}}{2}

N_{db}=Number of double bonds

N_{c} = Number of carbon atoms

N_{H} = Number of hydrogen atoms

N_{N} = Number of nitrogen atoms

The number of double bonds in the given formula - C_{4}H_{8}O

N_{db}= 4+1-\frac{8+0-0}{2}=1

The number of double bonds in the compound is one.

Therefore, probable structures is as follows.

(In attachment)

The structures I and III are ruled out from the probable structures because the signal in 13C-NMR appears at greater than 160 ppm.

alkene compounds I and II shows signal less than 140 ppm.

Hence, the probable structures III and IV are given as follows.

The carbonyl of structure I appear at 202 and ketone group of IV appears at 208 in 13C, which are greater than 160.

Hence, the molecular formula of the compound C_{4}H_{8}O having possible structure in which the signal appears at greater than 160 ppm are shown aw follows.

8 0
2 years ago
An atom of lead has a radius of and the average orbital speed of the electrons in it is about . Calculate the least possible unc
victus00 [196]

The question is incomplete. Here is the complete question.

An atom of lead has a radius of 154 pm and the average orbitalspeed of the electron in it is about 1.8x10^{8} m/s. Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of lead. Write your answer as a percentage of the average speed, and round it to significant 2 digits.

Answer: v% = 0.21 m/s

Explanation: To calculate the uncertainty, use <u>Heisenberg's Uncertainty Principle</u>, which states that:  ΔpΔx≥\frac{h}{4\pi }

where h is <u>Planck's constant</u> and it is equal to 6.626.10^{-34}m²kg/s.

Since p (momentum) is p = m.v:

mΔv.Δx ≥ \frac{h}{4\pi }

Δv = \frac{h}{4\pi.x.m }

Given that: r = x = 1.54.10^{-10}m and mass of an electron is m=9.1.10^{-31}kg

Δv = \frac{6.626.10^{-34} }{4.3.14.1.54.10^{-10}.9.1.10^{-31}}

Δv = 0.0376.10^{7}

As percentage of average speed:

Δv.\frac{1}{v}.100% = \frac{0.0376.10^{7} }{1.8.10^{8} }.10² = 0.021.10 = 0.21%

The least possible uncertainty in a speed of an electron is 0.21%.

5 0
2 years ago
What fraction of the carbon dioxide exhaled by animals is generated by the reactions of the citric acid cycle
puteri [66]

Answer:

1/3

Explanation:

Pyruvate is produced by the glycolysis in cytoplasm. The oxidation of pyruvate takes place in mitochondrial matrix.  

Pyruvate is converted to acetyl-CoA in the reaction given below:

Pyruvate + NAD⁺ + CoA-SH ⇒ acetyl-CoA + NADH + CO₂

1 molecule of carbon dioxide is eliminated from 1 molecule of pyruvate.

Also,  

2 molecules of carbon dioxide is eliminated from 2 molecules of pyruvate (as glucose on glycolysis yields 2 molecules of pyruvate).

Also, acetyl-CoA further goes into the citric acid cycle and produces 2 molecules of carbon dioxide.

Thus pyruvate produces total 3 molecules of CO₂  and hence glucose produces 6 molecules of CO₂ (as glucose on glycolysis yields 2 molecules of pyruvate)

Thus,

<u>Fraction = 2/6 = 1/3</u>

4 0
2 years ago
In which of these statements are protons, electrons, and neutrons correctly compared?
stiv31 [10]
<span>Quarks are present in protons and neutrons but not in electrons. Quarks are sub-atomic particles that have mass, but not an integer of charge. Protons and neutrons are made up of quarks, but electrons are not since they are energy travelling with a charge of energy, not matter with mass. Quarks have mass, therefore cannot be in electrons.</span>
5 0
2 years ago
Read 2 more answers
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