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Llana [10]
2 years ago
12

In the original mixtures, test tube 1 contained magnesium and a zinc compound, and test tube 2 contained zinc

Chemistry
1 answer:
FromTheMoon [43]2 years ago
5 0

Answer:

Test tube one is a compound because it is 2 elements combine..Test tube to is an element

Explanation:

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. The freezing point of an aqueous solution containing a nonelectrolyte solute is – 2.79 °C. What is the boiling point of this s
Semmy [17]

Answer:

Boiling point of the solution is 100.78°C

Explanation:

This is about colligative properties.

First of all, we need to calculate molality from the freezing point depression.

ΔT = Kf . m . i

As the solute is nonelectrolyte, i = 1

0°C - (-2.79°C) = 1.86 °C/m . m . 1

2.79°C / 1.86 m/°C = 1.5 m

Now, we go to the boiling point elevation

ΔT = Kb . m . i

Final T° - 100°C  =  0.52 °C/m . 1.5m . 1

Final T° =  0.52 °C/m . 1.5m . 1  + 100°C → 100.78°C

4 0
3 years ago
A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so
Xelga [282]

Answer : The equilibrium concentration of SCN^- in the trial solution is 4.58\times 10^{-8}M

Explanation :

First we have to calculate the initial moles of Fe^{3+} and SCN^-.

\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}

\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol

and,

\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}

\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol

The given balanced chemical reaction is,

Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)

Since 1 mole of Fe^{3+} reacts with 1 mole of SCN^- to give 1 mole of FeSCN^{2+}

The limiting reagent is, SCN^-

So, the number of moles of FeSCN^{2+} = 0.0020 mmole

Now we have to calculate the concentration of FeSCN^{2+}.

\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M

Using Beer-Lambert's law :

A=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution

l = path length

\epsilon = molar absorptivity coefficient

\epsilon and l are same for stock solution and dilute solution. So,

\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}

For trial solution:

The equilibrium concentration of SCN^- is,

[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]

[SCN^-]_{initial} = 0.00050 M

Now calculate the [FeSCN^{2+}].

C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M

Now calculate the concentration of SCN^-.

[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]

[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)

[SCN^-]_{eqm}=4.58\times 10^{-8}M

Therefore, the equilibrium concentration of SCN^- in the trial solution is 4.58\times 10^{-8}M

5 0
3 years ago
Calculate the standard entropy change for the following reaction cu(s) + 1/2 O2(g) —> cuo(s)
Evgen [1.6K]

3433Explanation:

the awser sucks and its not RIGHT

8 0
2 years ago
Read 2 more answers
HELP PLEASE<br>Explain why the elements of group 1 and 7 are quite reactive
Sauron [17]
Group 17 is the most readily reduced elements on the periodic table, meaning that they are so close to being a stable elements, only missing 1 electron to complete their valance electron shell. Thus they will essentially react with anything to get that last electron! 

Group 1 elements are extremely reactive because they are the most readily oxidized, they are very close to reaching stability by giving up only 1 electron. Thus they will react with almost anything to give up their electron. 
8 0
3 years ago
Please help
Snezhnost [94]
Relatively few hydrogen atoms
5 0
3 years ago
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