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Vlada [557]
3 years ago
8

HELP!!!!! i need help on this chemistry question

Chemistry
1 answer:
Alina [70]3 years ago
8 0

Answer: electronic configuration

Explanation:

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Number of grams in four moles of BaCO3
Irina-Kira [14]
This compound is also known as Barium Carbonate. 1 mole is equal to 1 moles BaCO3, or 197.3359<span> grams.</span>
5 0
2 years ago
A 20.0-milliliter sample of 0.200 M K2CO3 solution is added to 30.0 milliliters of 0.400 M Ba(NO3)2 solution.
jenyasd209 [6]

Answer:

(B) 0.160 M

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For K_2CO_3 :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of K_2CO_3 :

Moles=0.200 \times {20.0\times 10^{-3}}\ moles

Moles of K_2CO_3 = 0.004 moles

For Ba(NO_3)_2 :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume =30.0×10⁻³ L

Thus, moles of Ba(NO_3)_2 :

Moles=0.400\times {30.0\times 10^{-3}}\ moles

Moles of Ba(NO_3)_2  = 0.012 moles

According to the given reaction:

K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}

1 mole of potassium carbonate react with 1 mole of barium nitrate

0.004 moles potassium carbonate react with 0.004 mole of barium nitrate

Moles of barium nitrate  = 0.004 moles

Available moles of barium nitrate  =  0.012 moles

Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of potassium carbonate gives 1 mole of barium carbonate

Also,

0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate

Mole of barium carbonate = 0.004 moles

Also, consumed barium nitrate = 0.004 moles  (barium ions precipitate with carbonate ions)

Left over moles = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L

So, Concentration = 0.008/0.05 M = 0.160 M

<u>(B) is correct.</u>

4 0
3 years ago
How many moles of iron in a sample that contains 7.91×10^23 atoms of iron?
vazorg [7]

Answer: 1.31

Explanation:  

No.of moles = given no.of atoms/Avagadro number

= 7.91×10^23 / 6.022 x 10^23

= 1.31

therefore, no.of moles = 1.31

Hope it helped u,

pls mark as the brainliest

^_^

5 0
2 years ago
As you move from left to right across a period, what happens to the atomic radil?
egoroff_w [7]

Answer:

Moving Across a Period

Moving from left to right across a period, the atomic radius decreases. The nucleus of the atom gains protons moving from left to right, increasing the positive charge of the nucleus and increasing the attractive force of the nucleus upon the electrons

4 0
3 years ago
Purification of nickel can be achieved by electrorefining nickel from an impure nickel anode onto a pure nickel cathode in an el
Alexxandr [17]

Answer: 530 hours

Explanation:

The reduction of Nickel ions to nickel is shown as:

Ni^{2+}+2e^-\rightarrow Ni

96500\times 2=193000Coloumb of electricity deposits 1 mole of Nickel

1 mole of Nickel weighs = 58.7 g

Given quantity = 18.0 kg = 18000 g  (1kg=1000g)

58.7 g of Nickel is deposited by 193000 C of electricity

18000 g of Nickel is deposited by = \frac{193000}{58.7}\times 18000=59182282.8C of electricity

Q=I\times t

where Q= quantity of electricity in coloumbs  = 59182282.8C

I = current in amperes = 31.0 A

t= time in seconds = ?

59182282.8C=31.0A\times t

t=1909105.9sec

(1h=3600 sec)

t=530hours

Thus 530 hours are required to plate 18.0 kg of nickel onto the cathode if the current passed through the cell is held constant at 31.0 A

3 0
2 years ago
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