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Ksju [112]
3 years ago
6

the variable in an expierment that is being observed and changes in response to an independent variable

Chemistry
1 answer:
Oksanka [162]3 years ago
6 0
That is the dependent variable...
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When adenosine triphosphate (ATP) releases the energy it carries, it typically breaks apart into _____.
ddd [48]
Answer:

Adenosine diphosphate (ADP) and inorganic phosphate
8 0
2 years ago
How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --> __ AlCl3 + __
kiruha [24]

Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

  • determine the number of moles of HCl
  • determine the mole ratio,
  • use the mole ratio to calculate the number of moles of aluminum.
  • use RFM of Aluminum to determine the grams required.

<u>Moles </u><u>of </u><u>HCl</u>

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

x \: mol \:  =  \:  \frac{2 \:  \times  \: 35}{1000}  \\  = 0.07 \: moles \:

We have 0.07 moles of HCl.

<u>Mole </u><u>ratio</u>

  • Balanced equation

6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)

Hence mole ratio = 6 : 2 (HCl : Al

  • but moles of HCl is 0.07, therefore the moles of Al;

=  \frac{2}{6}  \times 0.07 \\  \:  = 0.0233333 \: moles

Therefore we have 0.0233333 moles of aluminum.

<u>Grams of </u><u>Aluminum</u>

We use the formula;

grams \:  = moles \:  \times  \: rfm

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

= 0.0233333 \: moles  \:  \times  \: 26.982 \:  \frac{g}{mol}  \\  = 0.625799 \: grams

The number of grams of aluminum required to react with HCl is 0.6258 g.

3 0
2 years ago
What chemical change is magnetizing steel
labwork [276]
Magnetizing steel is a physical change, not a chemical change.
4 0
3 years ago
If 23.6 g of hydrogen gas reacts with 28.3 g of nitrogen gas, what is the maximum amount of product that can be produced?
Aleonysh [2.5K]

Answer:

34.3 g NH3

Explanation:

M(H2) = 2*1 = 2 g/mol

M(N2) = 2*14 = 28 g/mol

M(NH3) = 14 + 3*1 = 17 g/mol

23.6 g H2* 1 mol/2 g = 11.8 mol H2

28.3 g N2 * 1 mol/28 g = 1.01 mol N2

                                 3H2 + N2 ------> 2NH3

from reaction         3 mol    1 mol

given                   11.8 mol    1.01 mol

We can see that H2 is given in excess, N2 is limiting reactant.

                                 3H2 + N2 ------> 2NH3

from reaction                     1 mol         2 mol

given                                 1.01 mol      x

x = 2*1.01/1= 2.02 mol NH3

2.02 mol * 17g/1 mol ≈ 34.3 g NH3

8 0
3 years ago
Find the mass of 4.5 moles of H3PO4
Contact [7]

Hey there!

H₃PO₄

Find molar mass.

H: 3 x 1.008 = 3.024

P: 1 x 30.97 = 30.97

O: 4 x 16 = 64

---------------------------------

                  97.994 grams

The mass of 1 mole of H₃PO₄ is 97.994 grams.

We have 4.5 moles.

97.994 x 4.5 = 440

The mass of 4.5 moles of H₃PO₄ is 440 grams.

Hope this helps!

6 0
3 years ago
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