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nikklg [1K]
3 years ago
10

!!!!HELP ASAP!!!! How does the scientific method help scientists?

Chemistry
2 answers:
denis-greek [22]3 years ago
7 0
B it helps then analyze their experience
photoshop1234 [79]3 years ago
5 0

Answer:

it helps them get reliable results

Explanation:

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Electromagnetic waves vary in
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I believe the answer is A. <span>wavelength and frequency. </span>
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PLEASE HURRY TIME LIMIT WILL GIVE BRAINLIEST A solution is made using 80.1 g of toluene (MM = 92.13 g/mol) and 80.0 g of benzene
Tom [10]
10.9 because of mass= moles/mv it’s just basic chemistry
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Two atoms that are isotopes of one another must have the same number of what?
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Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)
Dovator [93]

Answer:

1. 0.97 V

2. Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)

Explanation:

In this case, we can start with the <u>half-reactions</u>:

Ag^+~_(_a_q_)->~Ag_(_s_)

Al_(_s_)~->~Al^+^3~_(_a_q_)

With this in mind we can <u>add the electrons</u>:

Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)  <u>Reduction</u>

Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~ <u>Oxidation</u>

The reduction potential values for each half-reaction are:

Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_) - 0.69 V

Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_) -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:

Al_(_s_)~->~Al^+^3~+~2e^-~ +1.66 V

Finally, to calculate the overall potential we have to <u>add</u> the two values:

1.66 V - 0.69 V = <u>0.97 V</u>

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)

I hope it helps!

3 0
3 years ago
You find a compound composed only of element X and chlorine and you know that the compound is 13.10% X by mass. Each molecule of
8_murik_8 [283]

Answer:

Explanation:

So, the formula for the compound should be:

XCl_{6}

Now we assume that we have 1 mol of substance, so we can make calculations to know the molar mass of element X, as follows:

M_{Cl}=35.45g/mol\\

So we have that 6 moles weight 212.7g, and we can make a rule of three to know the weight of compound X:

212.7g\rightarrow 86.9\%\\x\rightarrow 13.1\%\\\\x=\frac{212.7g*13.1}{86.9} =32.06g

As we used 1 mol, we know that the molar mass is 32.06g/mol

So the element has a molar mass of 32.06 g/mol and an oxidation state of +6, with this information, we can assure that the element X is sulfur, so the compound is SCl_{6}

8 0
3 years ago
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