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3 years ago

9

Which term names how clouds form by the collection of water vapor in the atmosphere? A. evaporation B. condensation C. precipita

tion D. seepage

1 answer:

cluponka [151]3 years ago

5 0

The answer is B. Condensation

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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

3 years ago

How do ionic bonds differ from covalent bonds?

Alja [10]

Ionic bonds are a metal and a non metal bond and a covalent bond is two no metals banded together.

4 0

2 years ago

The following reaction is at equilibrium in a sealed container.

ale4655 [162]

Answer:

D.Lowering the temperature is the best option.

Explanation:

The value of equilibrium constants aren't changed with change in the pressure or concentrations of reactants and products in equilibrium. The only thing that changes the value of equilibrium constant is a change of temperature.

In the reaction below for example;

A + B <==>C+D

If you have moved the position of the equilibrium to the right (and so increased the amount of C and D), why hasn't the equilibrium constant increased?

Let's assume that the equilibrium constant mustn't change if you decrease the concentration of C - because equilibrium constants are constant at constant temperature. Why does the position of equilibrium move as it does?

If you decrease the concentration or pressure of C, the top of the Kc expression gets smaller. That would change the value of Kc. In order for that not to happen, the concentrations of C and D will have to increase again, and those of A and B must decrease. That happens until a new balance is reached when the value of the equilibrium constant expression reverts to what it was before.

3 0

3 years ago

A solid element that is malleable, a good conductor of electricity, and reacts with oxygen is classified as a(1) metal(2) metall

sukhopar [10]

Metalloid is the closest because noble gases are obviously out, nonmetals are terrible conductors, and metals are not malleable

6 0

3 years ago

Read 2 more answers

A certain gas is present in a 13.0 L cylinder at 1.0 atm pressure. If the pressure is increased to 2.0 atm , the volume of the g

Scorpion4ik [409]

<h3>Answer:</h3>

The gas obeys the Boyle's law

<h3>Explanation:</h3>

According to Boyle's law, the volume of a fixed mass of a gas and the pressure are inversely proportional at constant absolute temperature.
That is; $P\alpha\frac{1}{V}$
Therefore, $k=PV$ , where k is a constant
At varying volume and pressures while keeping absolute temperature constant; k = P1V1 =P2V2

In this case, we are given;

Initial Volume of 13.0 L at initial pressure of 1.0 atm

New volume of 6.5 L at new pressure of 2.0 atm

But, K = PV

Therefore,

k1 = P1V1

= 1.0 atm × 13.0 L

= 13 atm.L

k2 = P2V2

= 2.0 atm × 6.5 L

= 13 atm.L

Thus, k1=k2

Thus, the gas obeys the Boyle's law

8 0

3 years ago