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Nana76 [90]
3 years ago
15

ou discover that the complex decomposes in water. When 0.1000 g of the complex is dissolved in water with excess NaHg(SCN)4, all

of the Co(II) is precipitated as CoHg(SCN)4 (s). After the precipitate is washed and dried, its mass is found to be 0.1102 g. How many grams of cobalt are contained in the original 0.1000 grams of the complex
Chemistry
1 answer:
Eduardwww [97]3 years ago
7 0

Answer:

6.28x10⁻³ g

Explanation:

First we convert 0.1102 grams of CoHg(SCN)₄ into moles, using its <em>molar mass</em>:

  • 0.1102 g ÷ 491.9 g/mol = 2.24x10⁻⁴ mol CoHg(SCN)₄

There is 1 Co mol per CoHg(SCN)₄ mol, meaning there's also 2.24x10⁻⁴ moles of Co.

We now <u>convert 2.24x10⁻⁴ moles of Co into grams</u>, using its <em>molar mass</em>:

  • 2.24x10⁻⁴ mol Co * 28.01 g/mol = 6.28x10⁻³ g
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If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

6 0
3 years ago
What happens if you cross a heterozygous organism with a heterozygous organism?
maw [93]
If you cross a heterozygous organism with another heterozygous organism, pair two Tt and Tt. Below is a punnet square to show the result:

      T     t
T   TT   Tt
t    Tt     tt

This means that you have two heterozygous recessive and one homogenous dominant and one homogenous recessive. 
5 0
3 years ago
If 0.0714 moles of N2 gas occupies 1.25 L space, how many moles of N2 have a volume of 25.0 L? Assume temperature and pressure s
vfiekz [6]

Answer:

1.428 moles

Explanation:

If 0.0714 moles of N2 gas occupies 1.25 L space,

how many moles of N2 have a volume of 25.0 L?

Assume temperature and pressure stayed constant.

we experience it 0.0714 moles: 1.25L space

x moles : 25L of space

to get the x moles, cross multiply

(0.0714 x 25)/1.25

1.785/1.25 = 1.428 moles

8 0
3 years ago
A hydrogen-filled balloon was ignited and 2.10 g of hydrogen reacted with 16.8 g of oxygen. How many grams of water vapor were f
stiks02 [169]
Hydrogen + oxygen --> water
2,1g + 16,8g = x
x = 18,9g
5 0
3 years ago
A pharmaceutical company is making a large volume of nitrous oxide (NO). They predict they will be able to make a maximum amount
iragen [17]

Answer:

The answer is "Option B"

Explanation:

From the query, the following knowledge is derived:  

Yield in percentage = 47%  

Performance of theory = 4860 g  

Actual yield Rate =?  

The percentage return is defined simply by the ratio between both the real return as well as the conceptual return multiplied by the 100. It's also represented as numerically:

Rate = \frac{Existing \ Rate} {Theoretical \ Rate} \times 100

Now We can obtain the percent yield as followed using the above formula:  

\text{Yield in percentage}= \frac{Actual \ yield \ Rate} {Theorical \ Rate} \times  100

47\% = \frac{Actual \ yield \ Rate}{4860}

The value of the Actual yield Rate =47\% \times 4860

                                                        = \frac{47}{100} \times 4860 \\\\ = 2284.2 g

The Actual yield Rate= 2284.2 g.

3 0
2 years ago
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