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Minchanka [31]
3 years ago
6

If you heat the cup of soup in the microwave to make it boil. Will it have a higher total kinetic energy than the bowl of soup?

Yes or no
Chemistry
1 answer:
Paha777 [63]3 years ago
8 0

No, The kinetic energy depends upon the number of particles motion. The hot bowl of soup will have greater number of particles in motion and it will contain more kinetic energy than the cup of soup.

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If a book has a weight of 23.2 N on Earth, what is its mass?
Julli [10]

Answer:  Mass is 2,37 kg

Explanation: Weight G = mg, and g = 9.81 m/s² on Earth.

m = W/g = 23.2 N /  9.81 m/s²

8 0
2 years ago
What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


4 0
3 years ago
What is the volume of 8.80 g of CH4 gas at STP?
Ann [662]

Answer:

12.32 L.

Explanation:

The following data were obtained from the question:

Mass of CH4 = 8.80 g

Volume of CH4 =?

Next, we shall determine the number of mole in 8.80 g of CH4. This can be obtained as follow:

Mass of CH4 = 8.80 g

Molar mass of CH4 = 12 + (1×4) = 12 + 4 = 16 g/mol

Mole of CH4 =?

Mole = mass/Molar mass

Mole of CH4 = 8.80 / 16

Mole of CH4 = 0.55 mole.

Finally, we shall determine the volume of the gas at stp as illustrated below:

1 mole of a gas occupies 22.4 L at stp.

Therefore, 0.55 mole of CH4 will occupy = 0.55 × 22.4 = 12.32 L.

Thus, 8.80 g of CH4 occupies 12.32 L at STP.

6 0
3 years ago
g Which statement concerning the structure of the atom is correct? Protons and neutrons have most of the mass and occupy most of
qaws [65]

Protons and neutrons have most of the mass but occupy very little of the volume of the atom.

7 0
3 years ago
Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic
Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = 164.5\times 10^{-9} g

1 ng=10^{-9} g

Volume of the sample = V = 15.3 cm^3

Density of the lake water sample ,d= 1.00 g/cm^3

Mass of sample =  M = d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g

ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in 1 cm^3  of lake water = \frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g

Mass of arsenic in 0.710 km^3 lake water be m.

1 km^3=10^{15} cm^3

Mass of arsenic in 0.710\times 10^{15} cm^3 lake water :

m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

=7,633.66\times \frac{2.74}{41.90} days=499.19 days

1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

It will take 1.37 years to remove all of the arsenic from the lake.

3 0
3 years ago
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