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2 years ago

9

Two rocks one weighing 100m and the other weiging 200n are drpped from a 50-m cliff at the same time when both rocks are 10m fro

m the ground and air friction is ignored which is the same

1 answer:

Ratling [72]2 years ago

8 0

Answer:

Ignore all friction

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A sealed container can hold

svp [43]

0.216 moles of gas can the container hold if a sealed container can hold 0.325 L of gas at 1.00 atm and 293 K.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

R = gas constant = 0.08206 L.atm / mol K

T = temperature, Kelvin

V=5 L

P = 1.05 atm

T = 296 K

Putting value in the given equation:

$\frac{PV}{RT}=n$

$n= \frac{1.05 atm\; X \;5 L}{ 0.08206 L.atm / mol K X 296 K}$

Moles = 0.216 moles

Hence, 0.216 moles of gas can the container hold if a sealed container can hold 0.325 L of gas at 1.00 atm and 293 K.

Learn more about the ideal gas here:

brainly.com/question/27691721

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6 0

2 years ago

How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lo

riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

$(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)$

The expression used will be:

$\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = heat released by the reaction = ?

m = mass of benzene = 125 g

$c_{p,g}$ = specific heat of gaseous benzene = $1.06J/g^oC$

$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC$

$\Delta H_{vap}$ = enthalpy change for vaporization = $33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g$

Molar mass of benzene = 78.11 g/mole

Now put all the given values in the above expression, we get:

$\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]$

$\Delta H=-67682.5J=-67.7kJ$

Therefore, the energy removed must be, -67.7 kJ

4 0

2 years ago

How many electrons should be shown in the lewis symbol for hydrogen?

fredd [130]

999,956,000,000,000,000

6 0

3 years ago

Please answer!! I WILL GIVE EXTRA POINTS AND BRAINLIEST

sveta [45]

The answer is D I believe

5 0

3 years ago

Read 2 more answers

Carbon dioxide in the atmosphere dissolves in raindrops to produce carbonic acid (H2CO3), causing the pH of clean, unpolluted ra

anastassius [24]

Answer:

The range of [H⁺] is from 2.51 x 10⁻⁶ M to 6.31 x 10⁻⁶ M,

Explanation:

To answer this problem we need to keep in mind the <u>definition of pH</u>:

pH = -log [H⁺]

So now we <u>calculate [H⁺] using a pH value of 5.2 and of 5.6</u>:

5.2 = -log [H⁺]

-5.2 = log [H⁺]

$10^{-5.2}$ = [H⁺]

6.31 x 10⁻⁶ M = [H⁺]

5.6 = -log [H⁺]

-5.6 = log [H⁺]

$10^{-5.6}$ = [H⁺]

2.51 x 10⁻⁶ M = [H⁺]

6 0

2 years ago