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Aliun [14]
2 years ago
9

How many micrograms (ug) are in 3.4 x 10^-5 ounces (oz)

Chemistry
1 answer:
marysya [2.9K]2 years ago
8 0

Answer:

964ug

Explanation:

The problem here involves converting from one unit to another.

 We are to convert from ounces to micrograms.

                                    1ug  = 1 x 10⁻⁶g

                                    1oz  = 28.35g

       

So we first convert to grams from oz then take to ug:

 Solving:

                    1oz  = 28.35g

             3.4 x 10⁻⁵oz  will then give  3.4 x 10⁻⁵ x 28.35 = 9.64  x 10⁻⁴g

So;

                    1 x 10⁻⁶g    = 1ug

          9.64  x 10⁻⁴g will give \frac{9.64 x 10^{-4} }{1 x 10^{-6} }      = 9.64 x 10²ug or 964ug

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Answer:

(FeSCN⁺²) = 0.11 mM

Explanation:

Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2

M (Fe(NO₃)₃  = 0.200 M

V (Fe(NO₃)₃ =  10.63 mL

n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol

M (KSCN) =  0.00200 M

V (KSCN) = 1.42 mL

n (KSCN) =  0.00200 * 1.42 = 0.00284 mmol

Total volume = V (Fe(NO₃)₃  + V (KSCN)

                       = 10.63 + 1.42

                       = 12.05 mL

Limiting reactant = KSCN

So,

FeSCN⁺² = 0.00284 mmol

M (FeSCN⁺²) = 0.00284/12.05

                     = 0.000236 M

Excess reactant = (Fe(NO₃)₃

n(Fe(NO₃)₃ =  2.126 mmol -  0.00284 mmol

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For standard 2:

n (FeSCN⁺²) = 0.000236 * 4.63

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V(standard 2) = 4.63 + 5.17

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M (FeSCN⁺²)  = 0.00109/9.8

                      = 0.000111 M = 0.11 mM

Therefore, (FeSCN⁺²) = 0.11 mM

7 0
2 years ago
How many moles of potassium nitrate, KNO3 are present in a sample with a mass of 85.2 g?
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3 years ago
Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppo
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Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

CH_4+2O_2\rightarrow CO_2+2H_2O

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4

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left\ over=0g

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7 0
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When naming a binary compound, what ending do you use to represent anions?
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I believe You replace the ending of the elements name with -ide. example: magnesium flourine should should be magnesium flouride. 
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Answer:

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It forms ammonia gas.

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6 0
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