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11Alexandr11 [23.1K]
3 years ago
14

Atoms of element X weigh 32 times more than atoms of element Y. A compound has the formula: XY2 The ratio of the mass of X to th

e mass of Y in this compound equals: g
Chemistry
1 answer:
tamaranim1 [39]3 years ago
4 0

Answer:

16:1

Explanation:

Atoms of element X weigh 32 times more than atoms of element Y. We can write this in a symbolic way.

mX = 32 mY   [1]

where,

  • mX and mY are the masses of X and Y, respectively

A compound has the formula: XY₂, that is, in 1 molecule of XY₂ there is 1 atom of X and 2 atoms of Y. The ratio of the mass of X to the mass of Y in this compound equals:

mX/2 mY  [2]

If we substitute [1] in [2], we get:

mX/2 mY = 32 mY/2 mY = 16 = 16:1

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Answer: I believe it is critical mass

Explanation:

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Which pair of atoms has the highest electronegativity difference? Na-F Ca-F H-F C-F F-F
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The answer is Na-F. The F has highest electronegativity among these elements. So we need to find the element with smallest electronegativity. And this element is Na.
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4 years ago
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What is the mass, in grams, 6.5 mol of sodium nitrate (nano3)? please show all work?
ladessa [460]
Mass is the property of a physical body and the resistance to acceleration when a net force is applied on the body.  
The atomic mass of sodium (Na) is = 22.98 
The atomic mass of nitrate (N) is = 14.00 
The atomic mass of oxygen (O) is = 15.99  
The sodium nitrate (NaNO3) consists of the atomic masses of Na+N+(O)3 = 85 grams  
Therefore, the mass of 6.5 mol of sodium nitrate is = 6.5 * 1 mol of NaNO3 
 = 6.5 * (85) 
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6 0
3 years ago
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Cho m gam FeO tác dụng hết với dung dịch H2SO4, thu được 200 ml dung dịch FeSO4 1M. Giá trị của m là
BaLLatris [955]

<u>Answer:</u> The mass of FeO required is 14.37 g

<u>Explanation:</u>

Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Moles}}{\text{Volume}} ......(1)

We are given:  

Molarity of iron (II) sulfate = 1 M

Volume of solution = 200 mL = 0.200 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

\text{Moles of }FeSO_4=(1mol/L\times 0.200L)=0.200mol

The chemical equation for the reaction of FeO with sulfuric acid follows:

FeO+H_2SO_4\rightarrow FeSO_4+H_2O

By stoichiometry of the reaction:

If 1 mole of iron (II) sulfate is produced by 1 mole of FeO

So, 0.200 moles of iron (II) sulfate will produce = \frac{1}{1}\times 0.200=0.200mol of FeO

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We know, molar mass of FeO = 71.84 g/mol

Putting values in above equation, we get:

\text{Mass of }FeO=(0.200mol\times 71.84g/mol)=14.37g

Hence, the mass of FeO required is 14.37 g

4 0
3 years ago
A mixture of XO2 (P = 3.00 atm) and O2 (P = 1.00 atm) is placed in a container. This elementary reaction takes place at 27 °C: 2
sukhopar [10]

Answer:

a) \triangle G^{0} = 7.31 kJ/mol

b) K_{-1} = 0.0594 m^{-1} s^{-1}

Explanation:

Equation of reaction:

                                     2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)

Initial pressure                  3              1              0

Pressure change             2P           1P             2P

Total pressure = (3-2P) + (1-P) + (2P)

Total Pressure = 3.75 atm

(3-2P) + (1-P) + (2P) = 3.75

4 - P = 3.75

P = 4 - 3.75

P = 0.25 atm

Let us calculate the pressure of each of the components of the reaction:

Pressure of XO2 = 3 - 2P = 3 - 2(0.25)

Pressure of XO2 =2.5 atm

Pressure of O2 = 1 - P = 1 -0.25

Pressure of O2 = 0.75 atm

Pressure of XO3 = 2P = 2 * 0.25

Pressure of XO3 = 0.5 atm

From the reaction, equilibrium constant can be calculated using the formula:

K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }

K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}

Standard free energy:

\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol

b) value of k−1 at 27 °C, i.e. 300K

K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}

K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}

K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2}  }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}

6 0
3 years ago
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