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Mrrafil [7]
3 years ago
10

If a radioisotope has a half-life of 4 years, how much remains of a starting amount of 40 grams after 16 years have passed?

Chemistry
1 answer:
Gwar [14]3 years ago
5 0

Answer:

only 2.5 grams will be present after 16 years have passed

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How many anions are in 0.500 g of MgBr2
Oksana_A [137]

<u>Given:</u>

Mass of MgBr2 = 0.500 g

<u>To determine:</u>

Number of anions in 0.500 g MgBr2

<u>Explanation:</u>

Molar mass of MgBr2 = 24 + 2 (80) = 184 g/mol

Moles of MgBr2 = 0.500 g/184 g.mol-1 = 0.00271 moles

Based on stoichiometry-

1 mole of MgBr2 has 1 mole of Mg2+ cations and 2 moles of Br- anions

Therefore, 0.00271 moles of MgBr2 will have: 2 * 0.00271 = 0.00542 moles of Br-

Now,

1 mole of Br- contains 6.023 * 10²³ anions

0.00542 moles of Br- contain: 0.00542 * 6.023*10²³ = 3.264*10²¹ anions

Ans: There are 3.264*10²¹ anions in 0.5 g of MgBr2


4 0
3 years ago
At a certain temperature, 4.0 mol NH3 is introduced into a 2.0 L container, and the NH3 partially dissociates by the reaction. 2
xxMikexx [17]

Answer:

K = 3.37

Explanation:

2 NH₃(g) → N₂(g)  + 3H₂(g)

Initially we have 4 mol of ammonia, and in equilibrium we have 2 moles, so we have to think, that 2 moles have been reacted (4-2).

              2 NH₃(g)    →    N₂(g)  + 3H₂(g)

Initally       4moles             -            -

React        2moles           2m   +   3m

Eq             2 moles          2m        3m

We had produced 2 moles of nitrogen and 3 mol of H₂ (ratio is 2:3)

The expression for K is:  ( [H₂]³ . [N₂] ) / [NH₃]²

We have to divide the concentration /2L, cause we need MOLARITY to calculate K (mol/L)

K = ( (2m/2L) . (3m/2L)³ ) / (2m/2L)²

K = 27/8 / 1 → 3.37

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