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Alex
2 years ago
12

HElp me plzzzzz??????

Chemistry
1 answer:
iris [78.8K]2 years ago
4 0

Answer:

The answer to your question is given below

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

Zn + 2HCl —> ZnCl2 + H2

Thus, we can write out the atoms present in both the reactant and the product by doing a simple head count. The atoms present are listed below:

Element >>> Reactant >>> Product

Zn >>>>>>>> 1 >>>>>>>>>> 1

H >>>>>>>>> 2 >>>>>>>>> 2

Cl >>>>>>>>> 2 >>>>>>>>> 2

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A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

5 0
3 years ago
Pls help Would be much appreciated:)
Pie

Answer:

Ok so,  b. A redox reaction occurs in an electrochemical cell, where silver (Ag) is oxidized and nickel (Ni) is reduced - In voltaic cells, also called galvanic cells, oxidation occurs at the anode and reduction occurs at the cathode. A mnemonic for this is "An Ox. Red Cat." So since silver is oxidized, the silver half-cell is the anode. And the nickel half-cell is the cathode...

i. Write the half-reactions for this reaction, indicating the oxidation half-reaction and the reduction half-reaction- The substance having highest positive  potential will always get reduced and will undergo reduction reaction. Here, zinc will always undergo reduction reaction will get reduced

ii. Which metal is the anode, and which is the cathode?-The anode is where the oxidation reaction takes place. In other words, this is where the metal loses electrons. The cathode is where the reduction reaction takes place.

iii. Calculate the standard potential (voltage) of the cell

Look up the reduction potential,

E

⁰

red

, for the reduction half-reaction in a table of reduction potentials

Look up the reduction potential for the reverse of the oxidation half-reaction and reverse the sign to obtain the oxidation potential. For the oxidation half-reaction,

E

⁰

ox

=

-

E

⁰

red

.

iv. What kind of electrochemical cell is this? Explain your answer.

All parts in the electrochemical cells are labeled in second figure. Following are the part in electrochemical cells

1) Anode 2) Cathode 3) gold Stripe (Electrode) 4) Aluminium Glasses (Electrode) 5) Connecting wires 6) Battery

Explanation:

3 0
3 years ago
Science, please help and thank you!!
lapo4ka [179]
Swings hope this helps
8 0
3 years ago
A person's blood type is determined by the presence and absence of specific antigens on the cell membrane. Which of these would
sukhopar [10]

Answer:

C - no antibodies

Explanation:

I dont think there is any blood type without antibodies

5 0
3 years ago
What is the density of a substance that has a mass of 453g and a volume of 224mL?
Savatey [412]
453 divided by 224
density is roughly 2.02 g per ml
as a ml is 1 cm3 density is 2.02 grams per centimeter cubed
7 0
3 years ago
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