The movement of water from the atmosphere to the ground is called which of the following?

Chemistry

2 answers:

brilliants [131]3 years ago

7 0

D because it’s not going into then air or to the clouds

oee [108]3 years ago

6 0

The answer is C. Precipitation because we see precipitation as rain or snow or hail - from the atmosphere to the ground basically
Not A because the water already condensed to form clouds
Not B because water evaporates from the ocean or ground
Not D because plants transpire water into the atmosphere, which is a form of evaporation

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Which substance is the oxidizing agent in this reaction? fe2o3+3co→2fe+3co2 express your answer as a chemical formula?

erma4kov [3.2K]

Equation is as follow,

Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂

Oxidation:

3 CO → 3 CO₂

Oxidation state of C in CO is +2, and that in CO₂ is +4. So, carbon has lost 2 electrons per mole and 6 electrons per 3 moles hence,

3 CO → 3 CO₂ + 6 e⁻

Reduction:

Fe₂O₃ → 2 Fe

Oxidation state of Fe in Fe₂O₃ is +3 per atom, and that in Fe is 0. So, Iron has gained 3 electrons per atom and 6 electrons per 2 atoms hence,

Fe₂O₃ + 6e⁻ → 2 Fe

Result:
Iron in Fe₂O₃ has been reduced in this reaction and has played a role of oxidizing agent by oxidizing carbon from +2 state to +4 state.

4 0

3 years ago

How many grams of nan3 are required to produce 19.0 ft3 of nitrogen gas, about the size of an automotive air bag, if the gas has

Papessa [141]

The balanced chemical reaction is given as:

$2NaN_{3}(s)\rightarrow 2Na(s)+3N_{2}(g)$

Now, convert $19.0 ft^{3}$ into litres.

$1 ft^{3} = 28.3168$

So, $19.0 ft^{3} = 19\times 28.3168 = 538.0192 L$

Density is equal to the ratio of mass to the volume.

$D=\frac{M}{V}$

where, M = mass and V= volume $(538.0192 L)$

Substitute the value of density and volume in formula to get the value of mass.

$1.25 g/L=\frac{M}{538.0192 L}$

$1.25 g/L\times 538.0192 L= M$

$Mass = 672.524 g$

Now, number of moles of $N_{2}$ gas= $\frac{672.524 g}{28.02 g/mol}$

= $24.00 moles$

According to the reaction, 2 moles of sodium azide gives 3 moles of nitrogen gas.

Now, in 24.00 moles of nitrogen gas produced from= $\frac{2 moles of sodium azide}{3 moles of nitrogen gas}\times 24.00 moles$ of nitrogen gas, moles of sodium azide.