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Angelina_Jolie [31]

3 years ago

15

A student places 6.00 cups of water in a beaker. Two cups of water have the weight of 1.00 pounds. How much heat is needed to ra

ise the temperature of the water from 17.3°C to a boil? *
A) 15.9 J
B) 4.71 x 10⁵J
C) 470,849.28J
D) 9.42 x 10⁵ J

1 answer:

vivado [14]3 years ago

8 0

A

Explanation:

The awnser is a

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Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its total volume is 10 litres. How m

nlexa [21]

Answer:

The work done and heat absorbed are both -8,1 kJ

Explanation:

The work done in an isobaric process is defined as:

W = -P (Vf - Vi)

Where P is pressure ( 10 atm)

Vf = 10 L

Vi = 2 L

Thus, <em>W = -80 atm×L ≡ -8,1 kJ</em>

This is the work done in expansion of the gas. As the gas remains at the same temperature, there is no change in internal energy doing that all work was absorbed as heat.

I hope it helps!

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3 years ago

Explain the difference between rotation and revolution.

aleksandrvk [35]

Answer:

Rotation: spins on its axis

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3 years ago

Read 2 more answers

Suppose you have 75 gas-phase molecules of methanol (CH3OH) at T = 470 K. These molecules are contained in a spherical container

Katarina [22]

Answer:

The average pressure in the container due to these 75 gas molecules is $P=9.72 \times 10^{-16} Pa$

Explanation:

Here Pressure in a container is given as

$P=\frac{1}{3} \rho$

Here

P is the pressure which is to be calculated

ρ is the density of the gas which is to be calculated as below

$\rho =\frac{mass}{Volume of container}$

Here

mass is to be calculated for 75 gas phase molecules as

$m=n_{molecules} \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=75 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=3.98 \times 10^{-21} g$

Volume of container is 0.5 lts

So density is given as

$\rho =\frac{mass}{Volume of container}\\\rho =\frac{3.98 \times 10^{-21} \times 10^{-3} kg}{0.5 \times 10^{-3} m^3}\\\rho =7.97 \times 10^{-21}\, kg/m^3$

is the mean squared velocity which is given as

$=RMS^2$

Here RMS is the Root Mean Square speed given as 605 m/s so

$=RMS^2\\=(605)^2\\=366025$

Substituting the values in the equation and solving

$P=\frac{1}{3} \rho \\P=\frac{1}{3} \times 7.97 \times 10^{-21} \times 366025\\P=9.72 \times 10^{-16} Pa$

So the average pressure in the container due to these 75 gas molecules is $P=9.72 \times 10^{-16} Pa$

6 0

4 years ago

The decomposition of hydrogen peroxide follows first order kinetics and has a rate constant of 2.54 x 10-4 s-1 at a certain temp

pav-90 [236]

<u>Answer:</u> The initial concentration of hydrogen peroxide at the given temperature is 0.399 M

<u>Explanation:</u>

Decomposition of hydrogen peroxide is following first order kinetics.

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $2.54\times 10^{-4}s^{-1}$

t = time taken for decay process = 855 s

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 0.321 M

Putting values in above equation, we get:

$2.54\times 10^{-4}s^{-1}=\frac{2.303}{855s}\log \frac{[A_o]}{0.321}$

$[A_o]=0.399M$

Hence, the initial concentration of hydrogen peroxide at the given temperature is 0.399 M

7 0

3 years ago