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Alenkasestr [34]
1 year ago
8

Predict the product(s) and write a balanced equation for each of the following redox reactions:

Chemistry
1 answer:
Len [333]1 year ago
8 0

Mg(s) + HCL(aq) → MgCl 2(aq) + H2(g)

Magnesium added with hydrogen chloride which give Magnesium dichloride and hydrogen.

<h3>What is redox reaction ?</h3>

Redox reactions include a change in the oxidation state of the substrate. Loss of electrons or a rise in an element's oxidation state are both considered to be oxidation. Gaining electrons or lowering an object's or its atoms' oxidation state are both considered reductions. Combination, decomposition, displacement, and combustion reactions are the most frequent oxidation-reduction (redox) processes.

H2 + F2 = 2HF is the basic example of a redox reaction. Where hydrogen peroxide added with fluorine to form 2 molecule of hydrogen fluoride.

To know more about redox reaction visit :

brainly.com/question/1161855

#SPJ4

I understand the question you are looking for:

Predict the product(s) and write a balanced equation for each of the following redox reactions:

Mg(s) + 2HCL(aq)

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30,000 J of heat are added to 23.0 kg of steel to reach a final temperature of 140
kolbaska11 [484]

The initial temperature is 137.34 °C.

<u>Explanation:</u>

As the specific heat formula says that the heat energy required is directly proportional to the mass and change in temperature of the system.

Q = mcΔT

So, here the mass m is given as 23 kg, the specific heat of steel is given as c = 490 J/kg°C and the initial temperature is required to find with the final temperature being 140 °C. Also the heat energy required is 30,000 J.

ΔT =\frac{Q}{mc}

ΔT =\frac{30000}{23 \times 490} = \frac{30000}{11270} =2.66

Since the difference in temperature is 2.66, then the initial temperature will be

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140-Initial temperature = 2.66

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Thus, the initial temperature is 137.34 °C.

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Q2.  0.254 g of KHP (204 g/mol) titrated against 20.0 mL of unknown NaOH (40.0 g/mol) solution to get the end point of phenolpht
Vera_Pavlovna [14]

Answer:

<u>Mass concentration (g/L) </u><u><em>= 2.49g/L.</em></u>

Explanation:

No. of moles = \frac{mass}{molar mass}

= \frac{0.254}{204} = 0.001245 moles

Concentration of KHP (C1) in litres = n/v

= \frac{0.001245}{0.02} = 0.062 mol/L

We know that:

C_{1} V_{1} = C_{2} V_{2}

where c1v1 and c2v2 are the products of concentration and volumes of KHP and NaOH respectively.

Since mole ratio is 1 : 1.

1 mole of NaOH - 40g

0.001245 mole of NaOH = 40 × 0.001245 = 0.0498g

⇒0.0498g of NaOH was used during the titration

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<u><em>= 2.49g/L.</em></u>

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