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2 years ago

9

Zircon is a mineral with the empirical formula ZrSiO4. If all the zirconium is 90Zr, all the silicon is 28Si, and all the oxygen

is 16O, what mass of oxygen is present in 10. g of zircon?

1 answer:

Romashka-Z-Leto [24]2 years ago

4 0

Answer:

3.52g

Explanation:

Zr = 90 g/mol

Si = 28 g/mol

O = 16 g/mol

ZrSiO4 = 90 + 28 + (16* 4) = 182g/mol

Mass = Numberof moles * Molar mass

Mass = 1mol * 182g/mol = 182g

In one mole of ZrSiO4, there are 4 oxygen atoms, hence the mass is given as;

4 * 16 = 64g

Hence, 64g of oxygen is present in 182g of ZrSiO4.

10g would contain x

64 = 182

x = 10

x = ( 10 * 64 ) / 182

x = 640 / 182 = 3.52g

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maxonik [38]

Answer 1:
Equilibrium constant (K) mathematically expressed as the ratio of the concentration of products to concentration of reactant. In case of gaseous system, partial pressure is used, instead to concentration.

In present case, following reaction is involved:

2NO2 ↔ 2NO + O2

Here, K = $\frac{[PNO]^2[O2]}{[PNO2]^2}$

Given: At equilibrium, <span>PNO2= 0.247 atm, PNO = 0.0022atm, and PO2 = 0.0011 atm
</span>
Hence, K = $\frac{[0.0022]^2[0.0011]}{[0.247]^2}$
= 8.727 X 10^-8

Thus, equilibrium constant of reaction = 8.727 X 10^-8
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Answer 2:
Given: <span>PNO2= 0.192 atm, PNO = 0.021 atm, and PO2 = 0.037 atm.

Therefore, Reaction quotient = </span> $\frac{[PNO]^2[O2]}{[PNO2]^2}$
= $\frac{[0.021]^2[0.037]}{[0.192]^2}$
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Cacodyl, which has an intolerable garlicky odor and is used in the manufacture of cacodylic acid, a cotton herbicide, has a mass

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Answer:

The molecular formula of cacodyl is C₄H₁₂As₂.

Explanation:

<u>Let's assume we have 1 mol of cacodyl</u>, in that case we'd have 209.96 g of cacodyl and the<u> following masses of its components</u>:

209.96 g * 22.88/100 = 48.04 g C

209.96 g * 5.76/100 = 12.09 g H

209.96 g * 71.36/100 = 149.83 g As

Now we convert those masses into moles:

48.04 g C ÷ 12 g/mol = 4.00 mol C

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1 year ago

What is the hybridization of the central atom in each of the following? 1. Beryllium chloride 2. Nitrogen dioxide 3. Carbon tetr

Lina20 [59]

Answer :

(1) The hybridization of central atom beryllium in $BeCl_2$ is, $sp$

(2) The hybridization of central atom nitrogen in $NO_2$ is, $sp^2$

(3) The hybridization of central atom carbon in $CCl_4$ is, $sp^3$

(4) The hybridization of central atom xenon in $XeF_4$ is, $sp^3d^2$

Explanation :

Formula used :

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

Now we have to determine the hybridization of the following molecules.

(1) The given molecule is, $BeCl_2$

$\text{Number of electrons}=\frac{1}{2}\times [2+2]=2$

The number of electron pair are 2 that means the hybridization will be $sp$ and the electronic geometry of the molecule will be linear.

(2) The given molecule is, $NO_2$

$\text{Number of electrons}=\frac{1}{2}\times [4]=2$

If the sum of the number of sigma bonds, lone pair of electrons and odd electrons present is equal to three then the hybridization will be, $sp^2$ .

In nitrogen dioxide, there are two sigma bonds and one lone electron pair. So, the hybridization will be, $sp^2$ .

(3) The given molecule is, $CCl_4$

$\text{Number of electrons}=\frac{1}{2}\times [4+4]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

(4) The given molecule is, $XeF_4$

$\text{Number of electrons}=\frac{1}{2}\times [8+4]=6$

Bond pair electrons = 4

Lone pair electrons = 6 - 4 = 2

The number of electrons are 6 that means the hybridization will be $sp^3d^2$ and the electronic geometry of the molecule will be octahedral.

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Darina [25.2K]

The student should use the graduated cylinder. A graduated is the most common laboratory glassware when measuring volumes. It has calibrations by 1, 0.5 or 0.1 depending on the maximum volume. You have to make sure though, that you measure the volume by looking at the lower meniscus of the liquid at eye level.

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1 year ago

What is the yield of uranium from 2.50 kg U3O8?

mr_godi [17]

Answer: Thus the yield of uranium from 2.50 kg $U_3O_8$ is 2.12 kg

Explanation:

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To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$

moles of $U_3O_8=\frac{2.50\times 1000g}{842g/mol}=2.97mol$ (1kg=1000g)

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( 1kg=1000g)

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8 0

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