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2 years ago

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When a 3.22 g sample of an unknown hydrate of sodium sulfate, na2so4 ⋅ h2o(s), is heated, h2o (molar mass 18 g) is driven off. T

he mass of the anhydrous na2so4(s) (molar mass 142 g) that remains is 1.42 g. The value of x in the hydrate is?

1 answer:

Mariana [72]2 years ago

6 0

The value of X is 10 hence the formula of unknown hydrate sodium sulfate is NaSO4.10 H20

calculation

step 1:find the moles of NaSO4 and the moles of H2O

moles= mass/molar mass

moles of Na2SO4=1.42÷142=0.01 moles

moles of H20= mass of H2O/molar mass of H2O

mass of H2O= 3.22-1.42=1.8g

mole of H2O is therefore 1.8÷18=0.1 moles

step 2: find the mole ratio by dividing each mole by smallest number of mole (0.01)

that is Na2So4= 0.01/0.01 =1

H2O= 0.1/0.01=10

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The name "penicillin" is used for several closely-related antibiotics. A 1.2177 g sample of one of these compounds is burned, pr

masya89 [10]

<u>Answer:</u> The empirical formula for the given organic compound is $C_7H_{11}O_2SN$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$S_vN_wC_xH_yO_z+O_2\rightarrow CO_2+H_2O+SO_2+NO$

where, 'v', 'w' 'x', 'y' and 'z' are the subscripts of sulfur, nitrogen, carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=0.2829g$

Mass of $H_2O=0.1159g$

Mass of $SO_2=0.4503g$

Mass of NO = 0.2109 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

Molar mass of sulfur dioxide = 64 g/mol

Molar mass of nitrogen monoxide = 30 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 2.1654 g of carbon dioxide, $\frac{12}{44}\times 2.1654=0.590g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.6965 g of water, $\frac{2}{18}\times 0.6965=0.077g$ of hydrogen will be contained.

<u>For calculating the mass of sulfur:</u>

In 64 g of sulfur dioxide, 32 g of sulfur is contained.

So, in 0.4503 g of sulfur dioxide, $\frac{32}{64}\times 0.4503=0.225g$ of sulfur will be contained.

<u>For calculating the mass of nitrogen:</u>

In 30 g of nitrogen monoxide, 14 g of nitrogen is contained.

So, in 0.2109 g of nitrogen monoxide, $\frac{14}{30}\times 0.2109=0.098g$ of nitrogen will be contained.

Mass of oxygen in the compound = (1.2177) - (0.590 + 0.077 + 0.225 + 0.098) = 0.2277 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.590g}{12g/mole}=0.049moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.077g}{1g/mole}=0.077moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.2277g}{16g/mole}=0.0142moles$

Moles of Sulfur = $\frac{\text{Given mass of Sulfur}}{\text{Molar mass of sulfur}}=\frac{0.225g}{32g/mole}=0.007moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{0.098g}{14g/mole}=0.007moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.007 moles.

For Carbon = $\frac{0.049}{0.007}=7$

For Hydrogen = $\frac{0.077}{0.007}=11$

For Oxygen = $\frac{0.0142}{0.007}=2.03\approx 2$

For Sulfur = $\frac{0.007}{0.007}=1$

For Nitrogen = $\frac{0.007}{0.007}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O : S : N = 7 : 11 : 2 : 1 : 1

Hence, the empirical formula for the given compound is $C_7H_{11}O_2S_1N_1=C_7H_{11}O_2SN$

4 0

2 years ago

What scientific knowledge was necessary for Kroto, Smalley, and Curl to discover buckyballs? Check all that apply. All matter is

taurus [48]

All matter is made of atoms.

Atoms bond together in certain ways.

Carbon atoms have a specific mass.

8 0

2 years ago

Read 2 more answers

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N

Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

Now we have to calculate the dissociated concentration of $N_2O_4$ .

The balanced equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(aq)$

Initial conc. 1.0 M 0

At eqm. conc. (1.0-x) M (2x) M

As we are given,

The percent of dissociation of $N_2O_4$ = $\alpha$ = 28.0 %

So, the dissociate concentration of $N_2O_4$ = $C\alpha=1.0M\times \frac{28.0}{100}=0.28M$

The value of x = $C\alpha$ = 0.28 M

Now we have to calculate the concentration of $N_2O_4\text{ and }NO_2$ at equilibrium.

Concentration of $N_2O_4$ = 1.0 - x = 1.0 - 0.28 = 0.72 M

Concentration of $NO_2$ = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.56)^2}{0.72}=0.44$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.44

8 0

2 years ago

Which of the following combinations of particles represents an ion of net charge -1 and of mass number 82? Group of answer choic

noname [10]

Answer:

The correct choice is E (47 neutrons, 35 protons, 36 electrons)

Explanation:

A ion of net charge -1 means that the ion wins an e-.

We dismiss options B and C.

We also dismiss option A because neutrons + protons = 81.

neutrons + protons = mass number

So we have E and C.

We see that C has 47 neutrons, 36 protons and 36 electrons and, as the atomic number matches the number of protons, we find out on the Periodic Table that the atomic number 36 represents to Krypton, a noble gas. It is impossible! A noble gas never can't be a ion of net charge.

Option E is correct, It is an isotope of Br.

Br has 35 protons (its atomic number)

47 neutrons, that's why it is an Isotope (Br has always 45 neutrons)

As usual, Br has 35 electrons, if it represents a ion of net charge -1, it menas it won 1 e-.

5 0

2 years ago

For the reaction: MgF2(s) ⇌ Mg2+(aq) + 2F- (aq), Ksp= 6.4 × 10-9, the addition of 0.10 M NaF to the solution cause what effect o

galina1969 [7]

Answer:

Shifts the equilibrium to the left. reduces solubility.

Explanation:

MgF2(s) ↔ Mg2+(aq) + 2F-(aq)

S S 2S

∴ Ksp = 6.4 E-9 = [ Mg2+ ] * [ F- ]² = S * (2S)²

⇒ 4S² * S = 6.4 E-9

⇒ 4S³ = 6.4 E-9

⇒ S³ = 1.6 E-9

⇒ S = 1.1696 E-3 M

NaF(s) → Na+(aq) + F-(aq)

0.10M 0.10M 0.10M

MgF2(s) ↔ Mg2+(aq) + 2F-(aq)

S' S' 2S' + 0.10

⇒ Ksp = 6.4 E-9 = (S')*(2S' + 0.10)²

If we compare the concentration (0.10 M) of the ion with Ksp ( 6.4 E-9 ); thne we can neglect S' as adding:

⇒ 6.4 E-9 = (S')*(0.10)² = 0.01S'

⇒ S' = 6.4 E-7 M

∴ % S' = ( 6.4 E-7 / 0.1 )*100 = 6.4 E-4% <<< 5%, we can make the assumption

We can observe that S >> S' ( 1.1696 E-3 M >> 6.4 E-7 M ), which shows that the solubility is reduced by the efect of the common ion from the salt, which causes the equilibrium to shift to the left, precipitating part of MgF2(s).

8 0

2 years ago