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stiv31 [10]
3 years ago
7

In a redox reaction, one of the reactants must release

Chemistry
1 answer:
Leviafan [203]3 years ago
3 0
Redox reaction is the reduction and oxidation reaction. It is a chemical reaction that involves a transfer of electrons. It can mean loss of oxygen (oxidation) or gain of electrons (reduction). Oxydation is the process of where a sustance loses electrons, gains an oxygen atom/s, loses a hydrogen atom/s.  reduction is the opposite.
In a redox reaction, one of the reactants must release electrons.
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I have attached all the problems, but really if you just do one so I understand how to do it, that would be great!
Ivahew [28]

Answer:

Answer of question a is 345J.

Explanation:

In question a following is given in data:

-mass of iron (m) = 10.0 g

-temperature (ΔT) = final temperature- initial temperature= 100-25=  75 degree Celsius

-Specific Heat capacity of iron (c)= 0.46J/g°C.

Heat (Q)=?

Solution:

Formula for Heat is :

Q=m x c x ΔT

Q= 10 x 0.46 x 75

Q= 345 J.  

so, 345 joules of heat is needed to increase the temperature of 10 grams of iron.

  • From the above formula all other questions can easily be solved from the same procedure.
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3 years ago
Which phase is heat energy being released?
scZoUnD [109]

The answer is B
Vaporization
6 0
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Scientists frequently use all of the following to support the theory that organisms change over time except.
Pie

Answer:

4

Explanation:

8 0
3 years ago
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How to make science project on magnet​
Alex
You can put magnets on a fridge
4 0
3 years ago
A sample of 0.3283 g of an ionic compound containing the bromide ion (Br−) is dissolved in water and treated with an excess of A
Pavel [41]

Answer:

92.49 %

Explanation:

We first calculate the number of moles n of AgBr in 0.7127 g

n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g

n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol

Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and

From n = m/M

m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol

m = 0.0038 mol × 79.904 g/mol = 0.3036 g

% Br in compound = m₁/m₂ × 100%

m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)

m₂ = mass of compound = 0.3283 g

% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %

4 0
3 years ago
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