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4 years ago

Arrange the following compounds in order of decreasing basicity: MgO, Cs2O, Cl2O7, SnO2, P4O10

Chemistry

2 answers:

Natalka [10]4 years ago

5 0

Hey there!:

Genrally as the metalic character decreases basicity character decreases while acidic character increases .

Cs2O > MgO > SnO2 > P4O10 > Cl2O7

Answer 5

Hope this helps !

stich3 [128]4 years ago

5 0

Answer:

5) Cs2O > MgO > SnO2 > P4O10 > Cl2O7

Explanation:

Hello,

In this case, the decreasing basicity order depends on the metal contained into the oxide, in such a way, on the attached picture you will find how the basicity increases leftwards and downwards on the periodic table, as the metallic character does.

In this manner, as the cesium in the Cs₂O is the leftest atom, it is the most basic, next MgO, afterwards SnO₂, then P₄O₁₀ and finally Cl₂O₇, therefore, the order is: 5) Cs₂O> MgO > SnO₂> P₄O₁₀> Cl₂O₇.

Best regards.

You might be interested in

How many moles of PC15 can be produced from 51.0 g of Cl2 (and excess P4)?

slava [35]

Answer:

0.287 mole of PCl5.

Explanation:

We'll begin by calculating the number of mole in 51g of Cl2. This is illustrated below:

Molar mass of Cl2 = 2 x 35.5 = 71g/mol

Mass of Cl2 = 51g

Number of mole of Cl2 =..?

Mole = Mass /Molar Mass

Number of mole of Cl2 = 51/71 = 0.718 mole

Next, we shall write the balanced equation for the reaction. This is given below:

P4 + 10Cl2 → 4PCl5

Finally, we determine the number of mole of PCl5 produced from the reaction as follow:

From the balanced equation above,

10 moles of Cl2 reacted to produce 4 moles of PCl5.

Therefore, 0.718 mole of Cl2 will react to produce = (0.718 x 4)/10 = 0.287 mole of PCl5.

Therefore, 0.287 mole of PCl5 is produced from the reaction.

8 0

3 years ago

Analysis of an unknown sample indicated the sample contained 0.140 grams of N and 0.320 grams of O. The molecular mass of the co

vlada-n [284]

Answer:

the molecular formula of the compound is N2O4

Explanation:

- Find the empirical formula

mole of N present = mass of N divided by molar mass of N = 0.140/14 = 0.01 mole

mole of O present = mass of O divided by molar mass of O = 0.320/16 = 0.02 mole

Divide both by the smallest number of mole to determine the coefficient of each, the smallest number of mole is 0.01 thus:

quantity of N = 0.01/0.01 = 1

quantity of O = 0.02/0.01= 2

thus the empirical formula = NO2

- Now determine the molecular formula by finding the ratio of molecular formula and empirical formula

Molar mass of molecular formula = 92.02 amu = 92.02 g/mole

Molar mass of empirical formula NO2 = (14 + (16 x 2)) = 46 g/mole

the x factor = 92.02/46 = 2

Molecular formula = 2 x NO2 = N2O4

5 0

4 years ago

Given the following two quantities: 0.50 mol of CH4 and 1.0 mol of HCl,

Vinil7 [7]

Answer:

(a) HCl

(b) HCl

(d) HCl

Explanation:

<em>Given: </em>0.50 mol of CH₄ and 1.0 mol of HCl

Using stoichiometry we can calculate the answers to parts a, b, c, and d.

# of moles × Avogadro's number = # of atoms or molecules

Avogadro's number: 6.02 * 10²³

$\displaystyle 0.50\ \text{mol CH}_4 \cdot \frac{6.02\cdot 10^2^3 \ \text{atoms CH}_4}{1 \ \text{mol CH}_4} = 3.01 \cdot 10^2^3 \ \text{atoms CH}_4$
$\displaystyle 1.0\ \text{mol HCl} \cdot \frac{6.02\cdot 10^2^3 \ \text{atoms HCl}}{1 \ \text{mol HCl}} = 6.02 \cdot 10^2^3 \ \text{atoms HCl}$

HCl has more atoms than CH₄.

This is calculated the same way as Part (a); HCl has more molecules than CH₄.

Molar mass of CH₄ = 16.04 g/mol

Molar mass of HCl = 36.458 g/mol

$\displaystyle 0.50\ \text{mol CH}_4 \cdot \frac{16.04 \ \text{g CH}_4}{1 \ \text{mol CH}_4} = 8.02 \ \text{g CH}_4$
$\displaystyle 1.0\ \text{mol HCl} \cdot \frac{36.458 \ \text{g HCl}}{1 \ \text{mol HCl}} = 36.458 \ \text{g HCl}$

HCl has a greater mass than CH₄.

Assuming STP:

Molar volume of any gas at STP is 22.4 L/mol.

$\displaystyle 0.50\ \text{mol CH}_4 \cdot \frac{22.4 \ \text{L CH}_4}{1 \ \text{mol CH}_4} = 11.2 \ \text{L CH}_4$
$\displaystyle 1.0\ \text{mol HCl} \cdot \frac{22.4 \ \text{L HCl}}{1 \ \text{mol HCl}} = 22.4 \ \text{L HCl}$