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Usimov [2.4K]
3 years ago
12

The process by which cells chemically break down small food molecules and release chemical energy for the cell’s use is called

Chemistry
2 answers:
blagie [28]3 years ago
7 0

Answer:

It is called cellular respiration I guess!

Crazy boy [7]3 years ago
5 0

Answer: HEY THERE!

Explanation: HERE IS YOUR ANSWER

Cellular respiration is what happens inside cells when they use oxygen to transfer energy from food to ATP. Cellular respiration is essential to the transfer of matter and energy through living systems

<em>HOPE THIS HELPED AND HAVE A NICE DAY!</em>

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A student has a piece of aluminum metal. Which is the most reasonable assumption the student could make about the metal?
igor_vitrenko [27]

Aluminium belongs to 13th group of periodic table. It undergoes oxidation to given Al^+3 .

It is observed that when aluminium is added to a solution of copper sulphate the colour of the solution changes from blue to grey. It is due to formation of grey coloured solution of aluminium sulphate as

2Al^+3  + 3SO4^-2  ---> Al2(SO4)3


3 0
3 years ago
Read 2 more answers
A 150.0 mL solution of 2.888 M strontium nitrate is mixed with 200.0 mL of a 3.076 M sodium fluoride solution. Calculate the mas
Lelechka [254]

Answer:

Mass SrF2 produced = 38.63 g SrF2 produced

[Na^+]:  = 1.758 M

[NO3^-]:  = 1.238 M

[Sr^2+] = 0.3589 M

[F^-] = 2.36*10^-5 M

Explanation:

Step 1: Data given

Volume of 2.888M strontium nitrate = 150.0 mL = 0.150 L

Volume of 3.076 M sodium fluoride = 200.0 mL = 0.200 L

Step 2 : The balanced equation

Sr(NO3)2(aq) + 2NaF(aq) → SrF2(s) + 2NaNO3(aq) → Sr2+ + 2F- + 2

Step 3: Calculate moles strontium nitrate

Moles Sr(NO3)2 = Molarity * volume  

Moles Sr(NO3)2 = 2.888 M * 0.150 L

Moles Sr(NO3)2 = 0.4332 moles

Step 4: Calculate moles NaF

Moles NaF = 3.076 M * 0.200 L

Moles NaF = 0.6152 moles

It takes 2 moles F^- to precipitate 1 mole Sr^2+, so F^- is limiting.

Step 5: Calculate limiting reactant

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

NaF is the limiting reactant. It will completely be consumed (0.6152 moles).

Sr(NO3)2 is in excess. There will react 0.6152/2 = 0.3076 moles

Moles Sr^2+ precipitated by F^- = 0.3076

There will remain 0.4332 - 0.3076 = 0.1256 moles of Sr(NO3)2

Moles Sr^2+ no precipitated (left over) = 0.1256 moles

Step 6: Calculate moles SrF2  

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

For 0.6152 moles NaF we have 0.6152/2 = 0.3076 moles of SrF2

Mass SrF2 produced:  0.3076 mol * 125.6 g/mol = 38.63 g SrF2 produced

Step 7: Calculate concentration of [Na+] and [NO3-]

Since both Na^+ and NO3^- are spectator ions, and the final volume is 150 ml + 200 ml = 350 ml (0.350 L), the concentrations of Na^+ and NO3^- can be calculated as follows:

[Na^+]:  (200 ml)(3.076 M) = (350 ml)(x M) and x = 1.758 M

[NO3^-]:  (150 ml)(2.888 M)(2) = (350 ml)(x M) = 1.238 M

Step 8: Calculate [Sr^2+] and [F^-]

[Sr^2+] = 0.1256 moles/0.350 L = 0.3589 M

To find [F^-], one needs the Ksp for SrF2.  There are several values listed in the literature. I am using a value of 2x10^-10.

SrF2(s) <==> Sr^2+(aq) + 2F^-(aq)

Ksp = [Sr^2+][F^-]²

2x10^-10 = (0.3589)(x)²

x² = 5.57*10^-10

x = [F^-] = 2.36*10^-5 M

4 0
3 years ago
Part
r-ruslan [8.4K]

The molar concentration will be greater than 0.01 M KIO_{3}.


Since more of the compound was measured out than what was calculated, you can think of the solution as being 'stronger' than what it was calculated to be. Since a 'stronger' concentration results in a number that is higher, the molarity of this solution is going to be greater than 0.01 M.

7 0
3 years ago
Please help:/ I can’t figure out the answer
jekas [21]
Because other people cannot see what you are sending to somebody
7 0
3 years ago
Which chemical equation demonstrates the law of conservation of mass!
zlopas [31]

Answer:

F. 2NO + 02 —> 2NO

H. 4NH3 + 502 —> 4NO + 6H20

Explanation:

The law of conservation of mass states that matter can neither be created nor destroyed during a chemical reaction but can be convert from one form to another.

2NO + 02 —> 2NO

From the above, the total number of N on the left balance the total number on the right i.e 2 atoms of N on both side of the equation.

The total number of O on the left balance the total number on the right i.e 2 atoms of O on both side of the equation. This is certified by the law of conservation of mass.

4NH3 + 502 —> 4NO + 6H20

From the above, the total number of N on the left balance the total number on the right i.e 4 atoms of N on both side of the equation.

The total number of O on the left balance the total number on the right i.e 10 atoms of O on both side of the equation.

The total number of H on the left balance the total number on the right i.e 12 atoms of O on both side of the equation.

This is certified by the law of conservation of mass.

The rest equation did not conform to the law of conservation of mass as the atoms on the left side did not balance those on the right side

5 0
3 years ago
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