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Lostsunrise [7]
4 years ago
13

How do you answer this chem equation: 836.8J=14.5g*Cm*64

Chemistry
1 answer:
Alja [10]4 years ago
8 0
Is that a thing? Maybe you've said it wrong. Otherwise, I don't think you don't learn this in middle school.

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Which of the following atoms is MOST likely to lose an electron?<br> Li<br> F<br> C<br> Rb
saveliy_v [14]

The atom that is most likely to lose an electron is lithium (Li).

<h3>What is electronegativity?</h3>

Electronegativity is the tendency, or a measure of the ability, of an atom or molecule to attract electrons when forming bonds.

On the other hand, electropositivity is the tendency of an atom to release electrons to form a chemical bond.

Chemical elements that lose electrons become positively charged while elements that gain electrons become negatively charged.

Metals are most likely to lose electrons to form positive ions. Examples of metals are lithium in group 1, calcium in group 2, aluminium in group 3 etc.

Learn more about electropositivity at: brainly.com/question/17762711

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8 0
1 year ago
What is the percent yield if 4.65 g of copper?
Scilla [17]
What is the percent yield if 4.65 g of copper is produced when 1.87 g of aluminum reacts with excess copper (II) sulfate? 2Al (s) + 3CuSO4 (aq) → Al2 (SO4)3 (aq) + 3Cu
5 0
4 years ago
Can someone please help me
Margarita [4]

I think it is the first choice

Hope that helped!

5 0
3 years ago
Read 2 more answers
Please hurry this is due tomorrow morning
Llana [10]
Answer: G Atom 1 and Atom 4.. hope this helps and good luck!
3 0
3 years ago
The half-life of cobalt-60 is 5. 20 yr. how many milligrams of a 2. 000 mg sample remain after 6. 55 years?
Stels [109]

0.84 milligrams of a 2. 000 mg sample remain after 6. 55 years, according to radioactive decay.

Given data,

t\frac{1}{2} of Co-60 = 5.20years

amount of sample = 2.000mg initially = 0.002grams

According to radioactive decay,

N_{t} = N_{0}e^{-λt}

(N_{0} - 0.002 )λ = \frac{0.693}{t\frac{1}{2} } = \frac{0.693}{5.20}  = 0.133

According to radioactive decay,

N_{t} = N_{0}e^{-λt}

lnN_{t}  = lnN_{0} - λt

lnN_{t} = ln0.002 - (0.133×6.55)

       = -6.21 - 0.87 = -7.08 = 0.00084g = 0.84mg

Therefore, 0.84 milligrams of a 2. 000 mg sample remain after 6. 55 years.

Learn more about radioactive decay here:

brainly.com/question/1770619

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5 0
2 years ago
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