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valkas [14]
2 years ago
13

1. How does cellular respiration add carbon to the atmosphere?

Chemistry
1 answer:
asambeis [7]2 years ago
8 0
Glucose and oxygen are changed into energy and carbon dioxide during cellular respiration and that’s how carbon dioxide is released in the air.
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The gain of electrons by an element or ion in a reaction is called __.
dsp73

Answer:

Reduction

Explanation:

took the test

5 0
2 years ago
Sodium nitrate and lead (ii) acetate express your answer as a chemical equation. identify all of the phases in your answer. ente
ICE Princess25 [194]

Answer: 2NaNO_3(aq)+(CH_3COO)_2Pb(aq)\rightarrow 2CH_3COONa(aq)+Pb(NO_3)_2(aq)

Explanation: A double displacement reaction is one in which exchange of ions take place.

The compounds which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

Thus the exchange of ions take place and all the compounds are soluble so the chemical formulas are followed by the symbol (aq).

3 0
3 years ago
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How many L are in 1,500cm^3<br><img src="https://tex.z-dn.net/?f=1500%7Bcm%7D%5E%7B3%7D%20%3D%20%5C%3A%20...%20%20%5C%3A%20liter
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1.5 liters are in 1,500cm^3.
8 0
3 years ago
A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.08 g potassium hydrogen phthalate (KHC8H4O4
natka813 [3]

Answer:

A. 0.143 M

B. 0.0523 M

Explanation:

A.

Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).

KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄

The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:

1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol

The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.

5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:

M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M

B.

Let's consider the neutralization of potassium hydroxide and perchloric acid.

KOH + HClO₄ → KClO₄ + H₂O

When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.

M_{A} \times V_{A} = M_{B} \times V_{B}\\M_{A} = \frac{M_{B} \times V_{B}}{V_{A}} \\M_{A} = \frac{0.143 M \times 10.1mL}{27.6mL}\\M_{A} =0.0523 M

8 0
3 years ago
Hi, can someone help me with chemistry
Dominik [7]
Idk look it up on another website
6 0
3 years ago
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