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2 years ago

How do you balance redox equations in acidic solutions?

Chemistry

1 answer:

Anuta_ua [19.1K]2 years ago

7 0

Answer:

First, balance the half-reactions

Second, equalize the electrons

Third,add two reaction equations to get final answer

Explanation:

For example

H₂C₂0₄ + MnO⁻₄ ---------->CO₂+Mn²⁺

(i) Balancing the half reactions

H₂C₂O₄-------->2CO₂+2H⁺+2e⁻

5e⁻ +8H⁺+MnO₄⁻----------->Mn²⁺+4H₂O

(ii)

Equalizing the electrons

5H₂C₂O₄--------->10CO₂+10H⁺+10e⁻ ---here there is a factor of 5

10e⁻+16H⁺+2MnO₄⁻--------->2Mn²⁺+8H₂O -----here there is a factor of 2

(iii)

Add the two where electrons and some Hydrogen ions will cancel out

5H₂C₂O₄+6H⁺+2MnO₄⁻---->10CO₂+2Mn²⁺+8H₂O

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How is a gas's ability to fill a container different from that of a liquid or a solid?

svp [43]

Well a solid and a liquid is completely different from a gas. For a gas you need to have an air tight seal in order for the gas to stay in, or a special type of container so the gas doesn't decompose the container. And a liquid or a solid it's self explanatory. With some liquids you need special containers to hold the liquid so it doesn't decompose the container it's in.

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3 years ago

First person with the right answer gets brainliest thanks (btw the numbers on the right are the answers choose the right one)

kap26 [50]

Molar mass of FE2O3=2(55.85)+3(16)=159.7

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2 years ago

What is the measurement 1043. L rounded off to two significant figures?

Sloan [31]

Answer:

1000L

Explanation:

the 1 is a sig fig and since the 0 is between the 1 and 4 its also a significant number. to round them off you look at the 0,then look back at the 4 since its less than 5 u round down. then u replace the 43 with 0's

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3 years ago

Enter your answer in the provided box.

nexus9112 [7]

Answer:

V₂ = 50.93 L

Explanation:

Initial volume, $V_1=43.1\ L$

Initial temperature, $T_1=24^{\circ} C=24+273=297\ K$

Final temperature, $T_2=78^{\circ} C=78+273=351\ K$

We need to find the final volume of the gas. The relation between the volume and the temperature is given by :

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{43.1\times 351}{297}\\\\V_2=50.93\ L$

So, the final volume of the gas is 50.93 L.

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3 years ago

When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacit

pashok25 [27]

25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

<h3>Explanation</h3>

The process:

$\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq)$ .

How many moles of this process?

Relative atomic mass from a modern periodic table:

K: 39.098;
N: 14.007;
O: 15.999.

Molar mass of $\text{KNO}_3$ :

$M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}$ .

Number of moles of the process = Number of moles of $\text{KNO}_3$ dissolved:

$\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}$ .

What's the enthalpy change of this process?

$Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ}$ for $0.212162\;\text{mol}$ . By convention, the enthalpy change $\Delta H$ measures the energy change for each mole of a process.

$\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}$ .

The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.

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2 years ago