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Over [174]
3 years ago
9

What mass of znco3 contains 6.11×1022 o atoms? give your answer correctly to three significant digits?

Chemistry
2 answers:
nordsb [41]3 years ago
5 0

Using Avogadros number, we can get that 1 mole of an atom contain 6.022 x 10^23 atoms. Therefore we can use this conversion factor to get the number of moles:

moles ZnCO3 = 6.11 x 10^22 atoms * (1 mole / 6.022 x 10^23 atoms) = 0.10146 moles

 

The molar mass of ZnCO3 is about 125.39 g/mol, therefore the mass is:

mass ZnCO3 = 0.10146 moles * (125.39 g / mol)

<span>mass ZnCO3 = 12.72 g</span>

dexar [7]3 years ago
5 0

Answer is: mass of zinc carbonate is 4.241 grams.

N(O) = 6.11·10²²; number of oxygen atoms.

n(O) = N(O) ÷ Na (Avogadro constant).

n(O) = 6.11·10²² ÷ 6.022·10²³ 1/mol.

n(O) = 0.101 mol; amount of oxygen atoms.

One molecule of ZnCO₃ has three oxygen atoms: n(ZnCO₃) : n(O) = 1 : 3.

n(ZnCO₃) = 0.101 mol ÷ 3.

n(ZnCO₃) = 0.034 mol.

m(ZnCO₃) = n(ZnCO₃) · M(ZnCO₃).

m(ZnCO₃) = 0.034 mol · 125.39 g/mol.

m(ZnCO₃) = 4.241 g.

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5C + 6O2 = ? What will be the product of molecules formed from this equation?
harina [27]

Answer:5 moles ofCarbonmonoxide and 3.5 moles of oxygen gas.This in combine to yield carbon dioxide.

Explanation:

5C + 6O2----------5CO + 7/2O2.

When carbon combine with oxygen, carbon monoxide is formed first and it later recombine with oxygen to yield carbon dioxide.

6 0
3 years ago
Read 2 more answers
How many grams of hydrochloric acid will react completely with a block of gold is 3.2 cm by 3.8 cm by 2.8 cm, if the density of
JulijaS [17]
Answer:
mass of HCl = 243.5426 grams

Explanation:
1- we will get the mass of the reacting gold:
volume of gold = length * width * height
volume of gold = 3.2 * 3.8 * 2.8 = 34.048 cm^3 = 34.048 ml<span>
density = mass / volume
Therefore:
mass = density * volume
mass of gold = </span>19.3 * 34.048 = 657.1264 grams

2- we will get the number of moles of the reacting gold:
number of moles = mass / molar mass
number of moles = 657.1264 / 196.96657 
number of moles = 3.3362 moles

3- we will get the number of moles of the HCl:
First, we will balanced the given equation. The balanced equation will be as follows:
Au + 2HCl ......> AuCl2 + H2
This means that one mole of Au reacts with 2 moles of HCl.
Therefore 3.3362 moles will react with 2*3.3362 = 6.6724 moles of HCL

4- we will get the mass of the HCl:
From the periodic table:
molar mass of H = 1 gram
molar mass of Cl = 35.5 grams
Therefore:
molar mass of HCl = 1 + 35.5 = 36.5 grams/mole
number of moles = mass / molar mass
Therefore:
mass = number of moles * molar mass
mass of HCl = 6.6724 * 36.5
mass of HCl = 243.5426 grams

Hope this helps :)
4 0
3 years ago
4.3 moles of a gas are at a temperature of 28 degrees * C with a pressure of 1.631 atm. What volume does the gas occupy?
Shkiper50 [21]

Answer:

65.2L

Explanation:

Using the general gas equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (Litres)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (Kelvin)

According to the information provided in this question,

P = 1.631 atm

V = ?

n = 4.3 moles

T = 28°C = 28 + 273 = 301K

Using PV = nRT

V = nRT/P

V = 4.3 × 0.0821 × 301 ÷ 1.631

V = 106.26 ÷ 1.631

V = 65.15

Volume of the gas = 65.2L

7 0
3 years ago
In the absence of sodium methoxide, the same alkyl bromide gives a different product. Draw an arrowpushing mechanism to account
hoa [83]

Answer:

See explanation below

Explanation:

The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.

Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.

For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)

For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.

3 0
3 years ago
What is the correct chemical equation for the reaction between methane and oxygen to produce carbon dioxide and water?
kodGreya [7K]

Answer:

The answer to your question is:   CH₄     +    3/2 O₂     ⇒     CO₂    +   2 H₂O

Explanation:

Methane = CH₄

Oxygen = O

Carbon dioxide = CO₂

Water = H₂O

                      CH₄     +    3/2 O₂     ⇒     CO₂    +   2 H₂O

        This is the balanced equation

                 

3 0
3 years ago
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