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garik1379 [7]
3 years ago
14

1. Land-dwelling creatures depend on the the water cycle to ___________ fresh water from its salt solution in the Earth's oceans

to the atmosphere.
2. They also depend on the water cycle to ______________ water from the atmosphere down to the Earth's land.

Word Blank Bank:
Evaporate
Precipitate
Condensate
Infiltrate
Chemistry
2 answers:
Alex_Xolod [135]3 years ago
8 0
#1Evaporate and #2participate
Novosadov [1.4K]3 years ago
3 0
1 is evaporate and 2 is precipitate
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Hoochie [10]
First write out the eqn, and balance it.

next, take 12.8g/mr of aluminium
this will give u the mols of aluminium

next, take the ratio of HCL and Al, compare and find the mols of HCL.

take the mols of HCL/mr of HCL to give u mass

doneee
5 0
3 years ago
How many moles of O2 are produced when 0.425 mol of KO2 reacts in this fashion
Oksi-84 [34.3K]
The chemical reaction is expressed as:

<span> 4KO2+2CO2→2K2CO3+3O2 
</span>
We are given the amount of KO2 to be used in the reaction. This will be the starting point of our calculations. We do as follows:

0.425 mol KO2 ( 3 mol O2 / 4 mol KO2 ) = 0.319 mol O2 produced

Hope this answers the question. Have a nice day.
4 0
3 years ago
calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o
Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of Na_3PO_4 = 12.00 g

Mass of CaCl_2 = 10.0 g

Molar mass of Na_3PO_4 = 164 g/mol

Molar mass of CaCl_2 = 111 g/mol

Molar mass of NaCl = 58.5 g/mol

Molar mass of Ca_3(PO_4)_2 = 310 g/mol

First we have to calculate the moles of Na_3PO_4 and CaCl_2.

\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}

\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol

and,

\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}

\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2

From the balanced reaction we conclude that

As, 3 mole of CaCl_2 react with 2 mole of Na_3PO_4

So, 0.0901 moles of CaCl_2 react with \frac{2}{3}\times 0.0901=0.0601 moles of Na_3PO_4

From this we conclude that, Na_3PO_4 is an excess reagent because the given moles are greater than the required moles and CaCl_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaCl  and Ca_3(PO_4)_2

From the reaction, we conclude that

As, 3 mole of CaCl_2 react to give 6 mole of NaCl

So, 0.0901 mole of CaCl_2 react to give \frac{6}{3}\times 0.0901=0.1802 mole of NaCl

and,

As, 3 mole of CaCl_2 react to give 1 mole of Ca_3(PO_4)_2

So, 0.0901 mole of CaCl_2 react to give \frac{1}{3}\times 0.0901=0.030 mole of Ca_3(PO_4)_2

Now we have to calculate the mass of NaCl  and Ca_3(PO_4)_2

\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl

\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g

and,

\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2

\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g

Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

5 0
3 years ago
What does food provide to organisms?<br><br> A-homeostasis<br> B-shelter<br> C-energy<br> D-air
leva [86]
Energy would be the correct answer
5 0
2 years ago
Determine the oxidation number for nitrogenin
iris [78.8K]

Answer:

1(a) N = 3

(b) N = 0

(c) N = 5

(d) N = -2

(2) Molecular formula for benzene is C6H6

Explanation:

1(a) N02 1-

N + (2×-2) = -1

N-4 = -1

N = -1+4 = 3

(b) N2

2(N) = 0

N = 0/2 = 0

(c) NO2Cl

N + ( 2×-2) + (-1) = 0

N - 4 - 1 = 0

N - 5 = 0

N = 0+5 = 5

(d) N2H4

2(N) + (4×1) = 0

2N + 4 = 0

2N = 0 - 4 = -4

N = -4/2 = -2

(2) Molcular mass of benzene = 78g/mole = (6×12g of carbon) + (6×1g of hydrogen) = 72+6 = 78g/mole

Therefore, molecular formula for benzene is C6H6

7 0
3 years ago
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