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3 years ago

How does earth maintain its energy balance

1 answer:

5 0

<em>earth's energy balance describes how the incoming energy from the sun is used and returned to space. If incoming and outgoing energy are in balance, the earth's temperature remains constant.</em>

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PLEASE HELP EATHER ANSWER ONE OR ALL OR HOW MANY YOU WANT

lidiya [134]

Explanation:

Speed is the rate of an object moving along a path, whereas velocity is the direction of motion

Electricity is the flow of electric charge

3 0

1 year ago

Read 2 more answers

Calculating the Molecular Weight and Subunit Organization of a Protein From Its Metal Content The element molybdenum (atomic wei

Deffense [45]

Answer:

Dimer of two peptide chains with 1 mole of molybdenum metal each.

Explanation:

Percentage of molybdenum in protein = 0.08%

Molecular mass of nitrate reductase = 240,000 g

Mass of molybdenum = x

$0.08\%=\frac{x}{240,000 g}\times 100=192 g$

Moles of molybdenum = $\frac{192 g}{95.95 g/mol}=2.00 mol$

Each peptide chain of nitrate reductase contain 1 mole of molybdenum.

This means that nitrate reductase is composed of to two peptide chains. And in each peptide there is a single mole of molybdenum metal.

4 0

3 years ago

Covalent bonds result from the complete transfer of an electron from one atom to another. True or false

Fynjy0 [20]

False, covalent bonds result when atoms share electrons.

6 0

3 years ago

How strong are bonds between subatomic particles?

Aleks [24]

Answer:

they are indeed very strong

8 0

3 years ago

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