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nikdorinn [45]
3 years ago
10

How much of a 1.0 g polonium-214 sample remains after 818 microseconds? The half life of polonium-214 is 163.7 microseconds.​

Chemistry
1 answer:
Lera25 [3.4K]3 years ago
7 0
I need to get my stuff back in my room I don’t have a big deal
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18. The mass of a quantity of NiCl2 is 24.6 g. How many formula units are in the sample?
Charra [1.4K]

Answer is: A. 1.1 3 1023 NiCl2 formula units.

m(NiCl₂) = 24.6 g; mass of nickel(II) chloride.

M(NiCl₂) = 129.6 g/mol; molar mass of nickel(II) chloride.

n(NiCl₂) = m(NiCl₂) ÷ M(NiCl₂).

n(NiCl₂) = 24.6 g ÷ 129.6 g/mol.

n(NiCl₂) = 0.19 mol; amount of nickel(II) chloride.

Na = 6.022·10²³ 1/mol; Avogadro constant.

N(NiCl₂) = n(NiCl₂) · Na.

N(NiCl₂) = 0.19 mol · 6.022·10²³ 1/mol.

N(NiCl₂) = 1.13·10²³; number of formula units.

4 0
3 years ago
Read 2 more answers
Hemoglobin in your blood does not use elemental iron. It uses iron in the form of Fe2+(aq).
valina [46]

Explanation:

Balanced chemical reaction equation will be as follows.

     2Fe^{2+}(aq) + 2H^{+}(aq) \rightleftharpoons 2Fe^{3+}(aq) + H_{2}(g)

In human body, the neutral iron changes into Fe^{2+}(aq) cation. There will be an oxidation-half reaction and a reduction-half reaction. Equations for this reaction are as follows.

Oxidation: 2Fe^{2+}(aq) \rightleftharpoons 2Fe^{3+}(aq) + 2e^{-}[/tex] .... (1)

Reduction: 2H^{+}(aq) + 2e^{-} \rightleftharpoons H_{2}(g) ...... (2)

On adding both equation (1) and (2), the overall reaction equation will be as follows.

     2Fe^{2+}(aq) + 2H^{+}(aq) \rightleftharpoons 2Fe^{3+}(aq) + H_{2}(g)

Therefore, neutral iron is a part of Heme - b group of Hemoglobin and in an aqueous solution it dissolutes as a part of Heme group. Hence, then it becomes an Fe^{2+} cation.

3 0
3 years ago
a solution must be at a higher temperature than a pure solvent to boil. what colligative property can be employed to achieve thi
Kazeer [188]

Boiling-point elavation.

6 0
3 years ago
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How many grams of H2 are needed to react with 2.40 g of N2?
Georgia [21]
Is there an equation? I can't help if there's no equation involved.
5 0
3 years ago
A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o
julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

Number \, of \, moles, n  = \frac{Mass}{Molar \ mass} =  \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

3 0
3 years ago
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