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3 years ago

Help please !!!

Chemistry

1 answer:

VLD [36.1K]3 years ago

5 0

Answer:

imma have to say kinetic energy is transferred to the wheels 2 or b

Explanation:

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Water has a density of 1g/ml. what is the mass of the water if it fills a 10ml container?

notka56 [123]

It's 10.
Mass = density x volume
M = 1g/ml(10ml) = 10g

8 0

3 years ago

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134 grams of nitric acid is added to 512 grams of water. Calculate the molality of nitric acid.

Nana76 [90]

Answer: 4.15234 m

512 g H2O * $\frac{1 kg}{1000 g}$ = 0.512 kg H2O

Nitric Acid: HNO3 = 1.008 + 14.007 + 3(15.999) = 63.012 g/mol

H = 1.008 g/mol

N = 14.007 g/mol

O3 = 3*15.999

134 g HNO₃ * $\frac{mol}{63.012 g}$ = 2.126 mol

m = $\frac{2.126 mol}{0.512 kg}$ = 4.15234 m

6 0

3 years ago

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Which statement about carbon dioxide (CO2) is correct?

Zanzabum

Answer:

Explanation:

The carbon-oxygen double bond is polar.

However, carbon dioxide is a linear molecule. The two dipoles cancel out each other.

Thus, carbon dioxide is a non polar molecule with polar bond.

6 0

4 years ago

When the process of condensation occurs, the kinetic energy of particles

hammer [34]

Answer:

A. is insufficient to overcome intermolecular forces

Explanation:

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4 years ago

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A helium-filled balloon at 310.0 K and 1 atm, contains 0.05 g He, and has a volume of 1.21 L. It is placed in a freezer (T = 235

trapecia [35]

Answer : The value of $\Delta E$ of the gas is 2.79 Joules.

Explanation :

First we have to calculate the moles of helium.

$\text{Moles of helium}=\frac{\text{Mass of helium}}{\text{Molar mass of helium}}$

Molar mass of helium = 4 g/mole

$\text{Moles of helium}=\frac{0.05g}{4g/mole}=0.0125mole$

Now we have to calculate the heat.

Formula used :

$q=nc_p\Delta T\\\\q=nc_p(T_2-T_1)$

where,

q = heat

n = number of moles of helium gas = 0.0125 mole

$c_p$ = specific heat of helium = 20.8 J/mol.K

$T_1$ = initial temperature = 310.0 K

$T_2$ = final temperature = 235.0 K

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

$q=nc_p(T_2-T_1)$

$q=(0.0125mole)\times (20.8J/mol.K)\times (235.0-310.0)K$

$q=-19.5J$

Now we have top calculate the work done.

Formula used :

$w=-p\Delta V\\\\w=-p(V_2-V_1)$

where,

w = work done

p = pressure of the gas = 1 atm

$V_1$ = initial volume = 1.21 L

$V_2$ = final volume = 0.99 L

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

$w=-p(V_2-V_1)$

$w=-(1atm)\times (0.99-1.21)L$

$w=0.22L.artm=0.22\times 101.3J=22.29J$

conversion used : (1 L.atm = 101.3 J)

Now we have to calculate the value of $\Delta E$ of the gas.

$\Delta E=q+w$

$\Delta E=(-19.5J)+22.29J$

$\Delta E=2.79J$

Therefore, the value of $\Delta E$ of the gas is 2.79 Joules.

3 0

3 years ago