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3 years ago

What is the difference between the low tide and the high tide

Chemistry

1 answer:

9966 [12]3 years ago

6 0

The gravity pull of the moon is stronger when the tide is high and the pull is less when it is low tide. hope i helped

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At 35°C, K = 1.6 × 10^-5 for the reaction

TEA [102]

Answer:

a) [NOCl] = 0.968 M

[NO] = 0.032M

[Cl²] = 0.016M

b) [NOCl] = 1.992M

[NO] = 0.008 M

[Cl2] = 1.004 M

Explanation:

Step 1: Data given

Temperature = 35°C = 308K

K = 1.6 × 10^-5

Step 2: The reaction

2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)

For 2 moles NOCl we'll have 2 moles NO and 1 mol Cl2

Step 3

a. 2.0 mol pure NOCl in a 2.0 L flask

Concentration at the start:

Concentration = mol / volume

[NOCl] = mol / volume

[NOCl] = 2.0 / 2.0 L

[NOCl] = 1.0 M

[NO] = 0 M

[Cl] = 0M

Concentration at the equillibrium

[NOCl] = 1.0M - 2x

[NO] = 2x

[Cl2]= x

K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5

1.6*10^-5 = ((2x)² * x) / (1.0-2x)²

x = 0.016

[NOCl] = 1.0 - 2*0.016 = 0.968 M

[NO] = 2*0.016 = 0.032M

[Cl²] = 0.016M

b. 2.0 mol NOCl and 1.0 mol Cl2 in a 1.0 L flask

Concentration at the equillibrium

[NOCl] = 2.0 mol / 1.0 L = 2.0 M

[NO] = 0 M

[Cl2]= 1.0 mol / 1.0 L = 1.0 M

Concentration at the equillibrium

[NOCl] = 2.0M - 2x

[NO] = 2x

[Cl2]= 1.0 + x

K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5

1.6 *10^-5 = (2x)²*(1.0+x) / ((2.0-2x)²)

1.6 *10^-5= (2x)² * 1 )/2.0²

1.6 *10^-5= 4x² / 4 = x²

x = $\sqrt{1.6 *10^-5}$ = 4.0*10^-3

[NOCl] = 2.0 - 2*0.004 = 1.992M

[NO] = 2*0.004 = 0.008 M

[Cl2] = 1+ 0.004M = 1.004 M

5 0

3 years ago

A compound of formula XCl3 reacts with aqueous AgNO3 to yield solid AgCl according to the following equation: XCl3(aq)+3AgNO3(aq

Levart [38]

Answer:

Aluminum

Explanation:

$Moles =\frac {Given\ mass}{Molar\ mass}$

Mass of $AgCl$ = 1.68 g

Molar mass of $AgCl$ = 143.32 g/mol

Thus,

$Moles\ of\ AgCl=\frac {1.68}{169.87}=0.01172$

From the reaction below:

$XCl_3_{(aq)}+3AgNO_3_{(aq)}\rightarrow X(NO_3)_3_{(aq)}+3AgCl_{(s)}$

3 moles of $AgCl$ are produced when 1 mole of $XCl_3$ undergoes reaction.

So,

1 mole of $AgCl$ are produced when $\frac {1}{3}$ mole of $XCl_3$ undergoes reaction.

0.01172 mole of $AgCl$ are produced when $\frac {1}{3}\times 0.01172$ mole of $XCl_3$ undergoes reaction.

Thus, moles of $XCl_3$ = 0.0039 moles

Let the atomic mass of X = x g/mol

atomic mass of chlorine = 35.5 g/mol

Thus, Molar mass of $XCl_3$ = x + 3(35.5) g/mol = x + 106.5 g/mol

Moles = 0.0039 moles

Mass = 0.521 g

Thus, molar mass = Given mass/ Moles = 0.521 / 0.0039 = 133.5897 g/mol

So,

x + 106.5 = 133.5897

x = 27.0897 g/mol

This Atomic weight corresponds to Aluminum. Hence, X is aluminum.

8 0

4 years ago

Determine the empirical formula of a compound containing 48.38 grams of carbon, 8.12 grams of hydrogen, and 53.5 grams of oxygen

Dafna1 [17]

As with most stoichiometry problems, it is necessary to work in moles. The ratio of the moles of each element will provide the ratio of the atoms of each element.

Get the mass of each element by assuming a certain overall mass for the sample (100 g is a good mass to assume when working with percentages).

Remeber that percentages are a ratio multiplied by 100. You must convert percentages back to their decimal value before working with them.

(.4838) (100 g) = 48.38 g C

(.0812 ) (100 g) = 8.12 g H

(.5350) (100 g) = 53.38 g O

Convert the mass of each element to moles of each element using the atomic masses.

(48.38 g C) (1 mol/ 12.10 g C) = 4.028 mol C

(8.12 g H) (1 mol/ 1.008 g H) = 8.056 mol H

(53.38 g O) (1 mol/ 16.00 g O) = 3.336 mol O

Find the ratio or the moles of each element by dividing the number of moles of each by the smallest number of moles.

Use the mole ratio to write the empirical formula.

6 0

3 years ago

Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

BARSIC [14]

Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.