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3 years ago

The water cycle is balanced because _____.

2 answers:

8 0

I've seen this question before. I beleve the answer is:

the average annual precipitation on Earth equals the amount of water that evaporates

Inga [223]3 years ago

6 0

The answer is; the average annual precipitation on Earth equals the amount of water that evaporates

This water cycle is also known as the hydrological cycle. At any one moment in the cycle, there is water lost from the biosphere to the atmosphere through evapotranspiration. The water in the atmosphere also precipitates back to earth surface as rain or snow and the cycle continues.

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Really need help with this! Chemistry

Westkost [7]

Answer:

a) 0,5

Explanation:

If x=6 and y=2, then (2x-4y)/(x+y)=(2*6-4*2)/(6+2)=(12-8)/8=4/8= 0,5

5 0

2 years ago

Consider the reaction Ca(OH)2(s)→CaO(s)+H2O(l) with enthalpy of reaction ΔHrxn∘=65.2kJ/mol What is the enthalpy of formation of

Ganezh [65]

Answer:

ΔH °fCaO= -655.09 KJ/mol = ΔH °fCaO

Explanation:

ΔH °(reaction)a=ΔH °f(product)-ΔH °f(reactant)

where,

ΔH °f(product)=is the standard heat of formation of products

-ΔH °f(reactant)=is the standard heat of formation of reactants

The standard heat of formation of H 2 O ( l ) is − 285.8 k J / m o l .

The standard heat of formation of Ca(OH )2 is − 986.09 k J / m o l .

For the given reaction, the standard enthalpy change is calculated by the expression shown below.

ΔH °(reaction)a=ΔH °f(CaO+H2O)-ΔH °f(Ca(OH)₂

65.2 = ΔH °fCaO+( -265.8) -(-986.09)

= ΔH °fCaO-265.8+986.09

65.2 = ΔH °fCaO+ 720.29

65.2-720.29 = ΔH °fCaO

-655.09 KJ/mol == ΔH °fCaO

3 0

3 years ago

How many grams of a 1000-gram radioactive sample are left after 3 half-lives? *

faltersainse [42]

The answer is 125 gms.

3 0

4 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago

What do u do when you butter in a tub melts i put it in my fridge but after 1 hour it has some kind of water in it or something

schepotkina [342]

Answer: not sure either. not sure what exactly your asking because the wording is weird. (no offense)

6 0

3 years ago