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Given the data from the question, the mass of arsenic that contains 1.23×10²⁰ atoms is 0.0153 g
<h3>Avogadro's hypothesis </h3>
6.02×10²³ atoms = 1 mole of arsenic
But
1 mole of arsenic = 75 g
Thus, we can say that:
6.02×10²³ atoms = 75 g of arsenic
<h3>How to determine the mass that contains 1.23×10²⁰ atoms</h3>
6.02×10²³ atoms = 75 g of arsenic
Therefore,
1.23×10²⁰ atoms = (1.23×10²⁰ × 75) / 6.02×10²³ atoms)
1.23×10²⁰ atoms = 0.0153 g of arsenic
Thus, 1.23×10²⁰ atoms is present in 0.0153 g of arsenic
Learn more about Avogadro's number:
brainly.com/question/26141731
Answer:
An increase in the carbon dioxide concentration increases the rate at which carbon is incorporated into carbohydrate in the light-independent reaction, and so the rate of photosynthesis generally increases until limited by another factor.
Explanation:
Answer:
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Explanation:
<u>Step 1: </u>Data given
The solution contains 0.036 M Cu2+ and 0.044 M Fe2+
Ksp (CuS) = 1.3 × 10-36
Ksp (FeS) = 6.3 × 10-18
Step 2: Calculate precipitate
CuS → Cu^2+ + S^2- Ksp= 1.3*10^-36
FeS → Fe^2+ + S^2- Ksp= 6.3*10^-18
Calculate the minimum of amount needed to form precipitates:
Q=Ksp
<u>For copper</u> we have: Ksp=[Cu2+]*[S2-]
Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]
[S2-]= 3.61*10^-35 M
<u>For Iron</u> we have: Ksp=[Fe2+]*[S2-]
Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]
[S2-]= 1.43*10^-16 M
CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Answer:
false
Explanation:
As we know that in sodium-potassium pump .
sodium potassium move 3Na+ outside the cells
and moving 2k+ inside the cells
so that we can say that given statement is false
Answer FALSE
