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dolphi86 [110]
3 years ago
10

What is the number of cations is present in 1.17 g of sodium chloride? Given Na=6*10power 23 mol

Chemistry
2 answers:
SpyIntel [72]3 years ago
6 0

Answer:

1.21x10^{22}cationsNa^+

Explanation:

Hello,

In this case, one knows that sodium chloride's formula is NaCl, therefore, the amount of cations is computed via the following stoichiometric relationship, considering the Avogadro's number to relate moles of sodium cations with their actual amount:

1.17gNaCl*\frac{1molNaCl}{58.45gNaCl}*\frac{1molNa^+}{1molNaCl}*  \frac{6.022x10^{23}cationsNa^+}{1molNa^+} =1.21x10^{22}cationsNa^+

Best regards.

natita [175]3 years ago
5 0
1.17×6×1023 Wich equals 7.02×1023 Your final answer is 7.02×1023
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Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy for the following redox
stealth61 [152]

Answer:

-973 KJ

Explanation:

The balanced reaction equation is;

N2H4(aq) + 2Cl2(g) + 4OH^-(aq)---------> 4Cl-(aq) + 4H ^+(aq) + 4OH^-(aq) + N2(g)

Reduction potential of hydrazine = -1.16 V

Reduction potential of chlorine = 1.36 V

From;

E°cell= E°cathode - E°anode

E°cell= 1.36 - (-1.16)

E°cell= 2.52 V

∆G°=- nFE°cell

n= number of moles of electrons = 4

F= Faraday's constant = 96500 C

E°cell = 2.52 V

∆G°=- (4 × 96500 × 2.52)

∆G°= -972720 J

∆G°= -972.72 KJ

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4 years ago
Match each of the following topical dosage forms with its correct definition.
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Answer:

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8 0
2 years ago
Zachary adds 26.64 g to 12.557 g. How many significant figures should his answer have?
asambeis [7]

This answer should have four signifigant features

Explanation:

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5 0
4 years ago
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Perform the calculation and record the answer with the correct number of significant figures.
kotykmax [81]

The question is incomplete, here is the complete question:

A. (6.5-6.10)/3.19

B. (34.123 + 9.60) / (98.7654 - 9.249)

<u>Answer:</u>

<u>For A:</u> The answer becomes 0.1

<u>For B:</u> The answer becomes 0.4884

<u>Explanation:</u>

Significant figures are defined as the figures present in a number that expresses the magnitude of a quantity to a specific degree of accuracy.

Rules for the identification of significant figures:

  • Digits from 1 to 9 are always significant and have infinite number of significant figures.
  • All non-zero numbers are always significant. For example: 664, 6.64 and 66.4 all have three significant figures.
  • All zeros between the integers are always significant. For example: 5018, 5.018 and 50.18 all have four significant figures.
  • All zeros preceding the first integers are never significant. For example: 0.00058 has two significant figures.
  • All zeros after the decimal point are always significant. For example: 2.500, 25.00 and 250.0 all have four significant figures.
  • All zeroes used solely for spacing the decimal point are not significant. For example: 10000 has one significant figure.

<u>Rule applied for addition and subtraction:</u>

The least precise number present after the decimal point determines the number of significant figures in the answer.

<u>Rule applied for multiplication and division:</u>

In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits

  • <u>For A:</u> (6.5-6.10)/3.19

This a a problem of subtraction and division.

First, the subtraction is carried out.

\Rightarow \frac{6.5-6.10}{3.19}=\frac{0.4}{3.19}

Here, the least precise number after decimal was 1.

\Rightarrow \frac{0.4}{3.19}=0.125

Here, the least precise number of significant digit is 1. So, the answer becomes 0.1

  • <u>For B:</u> (34.123 + 9.60) / (98.7654 - 9.249)

This a a problem of subtraction, addition and division.

First, the subtraction and addition is carried out.

\Rightarow \frac{34.123+9.60}{98.7654-9.249}=\frac{43.723}{89.5164}=\frac{43.72}{89.516}

Here, the least precise number after decimal in addition are 2 and in subtraction are 3

\Rightarrow \frac{43.72}{89.516}=0.48840

Here, the least precise number of significant digit are 4. So, the answer becomes 0.4884

6 0
3 years ago
Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to
zhenek [66]

Answer:

<u>D. It will decrease by a factor of 4</u>

Explanation:

According to the question , the equation follows :

A+B\rightarrow C+D

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

Rate\ \alpha [A]^{a}[B]^{b}

r=[A]^{a}[B]^{b}.................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

r'=[A']^{a}[B']^{b}

Put the value of [A'] , [B'] and r' in the above equation:

2r=[2A]^{a}[B]^{b}...........(2)

Divide equation (1) by (2) we , get

\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}

2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}

Here A and A cancel each other

B and B cancel each other

We get,

2= 2^{a}\times 1^{b}

1^b = 1 ( power of 1 = 1)

2= 2^{a}

This is possible only when a = 1

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1 + b = 3

b =3 -1  = 2

b = 2

Hence the rate law becomes :

r=[A]^{a}[B]^{b}

<u>r=[A]^{1}[B]^{2}.............(3)</u>

Look in the question now, it is asked to calculate the concentration of [B],if  cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

r'=[A]^{1}[B']^{2}

r'=[A]^{1}(\frac{1}{2}[B])^{2}

r'=\frac{1}{4}[A]^{1}[B]^{2}............(a)

But

r=[A]^{a}[B]^{b}..............(b)

Compare equation (a) and (b) , we get

new rate r' =

<u>r' = 1/4 r</u>

7 0
3 years ago
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