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kati45 [8]
3 years ago
15

How many grams of vanadium would contain the same number of atoms as 58.693 g of nickel

Chemistry
1 answer:
Doss [256]3 years ago
4 0

To answer this question, we'll need the molar masses of nickel and vanadium:

Molar mass of nickel: 58.693g/mol

Molar mass of vanadium: 50.941g/mol

These values represent the mass of one mole of the element. A mole is the weight of Avogadro's number of atoms of an element (6.02 x 10^23 atoms).

We are given 58.693g of nickel, which is equivalent to one mole of nickel. Therefore, you will get the same number of atoms from one mole of vanadium, or 50.941g of vanadium.

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How many atoms of Mg are present in 97.22 grams of Mg?
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Answer:

2.439 x 10E24 atoms

Explanation:

from

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What volume of a 0.160 MM solution of KOHKOH must be added to 550.0 mLmL of the acidic solution to completely neutralize all of
puteri [66]

The given question is incomplete. The complete question is :

What volume of 0.160 M solution of KOH must be added to 550.0 mL of the acidic solution to completely neutralize all of the 0.150 M hydrochloric acid?

Answer: Volume in liters to three significant figures is 0.516 L

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH

We are given:

n_1=1\\M_1=0.50M\\V_1=550.0mL\\n_2=1\\M_2=0.160M\\V_2=?

Putting values in above equation, we get:

1\times 0.150\times 550.0=1\times 0.160\times V_2\\\\V_2=516mL=0.516L   (1L=1000ml)

Thus volume in liters to three significant figures is 0.516 L

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Where on the pH scale would you find acids? Bases? What is unique about a pH of 7?
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If you complete and balance the following oxidation-reduction reaction in basic solution NO2−(aq) + Al(s) → NH3(aq) + Al(OH)4−(a
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Answer:

NO2- + 5H2O + 2Al + OH- → NH3 + 2Al(OH)4-

There is 1 hydroxide ion, on the reactant side

Explanation:

NO2−(aq) + Al(s) → NH3(aq) + Al(OH)4−(aq)

Step 1: The half reactions

NO2- (aq) → NH3(g)

Al(s) → Al(OH)4-

<u>Step 2: </u>Balancing electrons

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On the left side N has an oxidation number of +3 and on the right side -3.

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On the left side, Al has an oxidation number of 0 and on the right side +3.

Al(s) → Al(OH)4- +3e-

To have the same amount of electrons transfered, we have to multiply the second reaction by 2

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<u>Step 3:</u> Balance with OH/H2O

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<u>Step 4:</u> The netto reaction

NO2- + 5H2O + 2Al + 8OH-  → NH3 +7OH- + 2Al(OH)4-

NO2- + 5H2O + 2Al + OH- → NH3 + 2Al(OH)4-

There is 1 hydroxide ion, on the reactant side

6 0
4 years ago
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