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Anna007 [38]
2 years ago
5

A sample of a compound contains only the elements sodium, sulfur, and oxygen. it is found by analysis to contain 0.979 g na, 1.3

65 g s, and 1.021 g o. determine its empirical formula.
Chemistry
2 answers:
Aloiza [94]2 years ago
4 0
To determine the empirical formula for the compound that contains <span>0.979 g Na, 1.365 g S, and 1.021 g O, we convert these to mole units. The molar masses to be used are:

Molar mass of Na = 23 g/mol
</span>Molar mass of S = 32 g/mol
Molar mass of O = 16 g/ mol

The number of moles is obtained using the molar mass for each element.

moles Na = 0.979 g Na/ 23 g/mol Na = 0.04256
moles S = 1.365 g Na/ 32 g/mol Na = 0.04265
moles O = 1.021 g O/ 16 g/mol Na = 0.06326

We then divide each with the smallest number of moles obtained. 

Na: 0.04256/ 0.04256 = 1 
S: 0.04265/ 0.04256 = 1.002 ≈ 1
O: 0.06326/ 0.04256 = 1.49 ≈ 1.5

We then have an empirical formula of NaSO₁.₅. However, chemical formulas must have only integers as subscripts, thus, we multiply each to 2. The empirical formula is then Na₂S₂O₃ also known as sodium thiosulfate.
Mumz [18]2 years ago
3 0
First, convert the masses into moles using the molar weights: 23 g Na/mol, 32.06 g S/mol and 16 g O/mol.

Mol Na: 0.979/23 = 0.04256522
Mol S: 1.365/32.06 = 0.04257642
Mol O: 1.021/16 = 0.0638125

The least value is mol Na. Thus, divide all mole compositions to that of mol Na:
Mol Na: 0.04256522/0.04256522 = 1
Mol S: 0.04257642/0.04256522 = 1
Mol O: 0.0638125/0.04256522 = 1.5

The ratios must be whole numbers. To make 1.5 a whole number, multiply all ratios with 2 to keep it uniform.
Mol Na: 1*2 = 2
Mol S: 1*2 = 2
Mol O: 1.5*2 = 3

<em>Thus, the empirical formula is Na₂S₂O₃</em>.
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Three samples of gas each exert 740 mm hg in separate 2 L containers what pressure do they exert if they are all placed in a sin
Nesterboy [21]

Answer:

P(total) = 1110 mmHg

Explanation:

According to the Dalton law of partial pressure,

The pressure exerted by mixture of gases are equal to the sum of partial pressure of individual gases.

P(total) = P1 + P2 + P3+ .....+ Pn

Given data:

Sample A = 740 mmHg

Sample B = 740 mmHg

Sample C = 740 mmHg

Total pressure = ?

Solution:

<em>Sample A:</em>

P₁V₁ = P₂V₂

P₂ = P₁V₁ / V₂

P₂ = 740 mmHg × 2L/4L

P₂ = 370 mmHg

<em>Sample B:</em>

P₁V₁ = P₂V₂

P₂ = P₁V₁ / V₂

P₂ = 740 mmHg × 2L/4L

P₂ = 370 mmHg

<em>Sample C:</em>

P₁V₁ = P₂V₂

P₂ = P₁V₁ / V₂

P₂ = 740 mmHg × 2L/4L

P₂ = 370 mmHg

Total pressure:

P(total) = P1 + P2 + P3

P(total) =  370 mmHg +  370 mmHg+  370 mmHg

P(total) = 1110 mmHg

6 0
2 years ago
A solution is prepared by adding 6.24 g of benzene (C 6H 6, 78.11 g/mol) to 80.74 g of cyclohexane (C 6H 12, 84.16 g/mol). Calcu
tekilochka [14]

Answer:

x_B=0.0769

m=0.990m

Explanation:

Hello,

In this case, we can compute the mole fraction of benzene by using the following formula:

x_B=\frac{n_B}{n_B+n_C}

Whereas n accounts for the moles of each substance, thus, we compute them by using molar mass of benzene and cyclohexane:

n_B=6.24g*\frac{1mol}{78.11g}=0.0799mol\\ \\n_C=80.74g*\frac{1mol}{84.16g} =0.959mol

Thus, we compute the mole fraction:

x_B=\frac{0.0799mol}{0.0799mol+0.959mol}\\ \\x_B=0.0769

Next, for the molality, we define it as:

m=\frac{n_B}{m_C}

Whereas we also use the moles of benzene but rather than the moles of cyclohexane, its mass in kilograms (0.08074 kg), thus, we obtain:

m=\frac{0.0799mol}{0.08074kg}=0.990mol/kg

Or just 0.990 m in molal units (mol/kg).

Best regards.

7 0
2 years ago
How many grams of NH3 can be prepared from 85.5 grams of N2 and 17.3 grams of H2 ?
Tcecarenko [31]
N2 + 3H2 ---> 2NH3

mass of N2 = 28g
mass of H2 = 2g
mass of NH3 = 17g

according to the reaction:
28g N2----------------- 3*2g H2
85,5g N2-------------------- x
x = 18,32g H2 >>>  so, nitrogen is excess

according to the reaction:
2*3g H2---------------------- 2*17g NH3
17,3g H2 ------------------------- x
x = 98,03g NH3

<u>answer: 98,03g of NH3</u>
4 0
2 years ago
The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW =
Varvara68 [4.7K]

Answer:

9.744g of monosodium succinate.

4.925g of disodium succinate.

Explanation:

To find pH of the buffer produced by the mixture of monosodium succinate-Disodium succinate is obtained from H-H equation:

pH = pKa + log ([Na₂Suc] / [NaHSuc])

As you want a pH of 5.28 and pKa is 5.64:

5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc])

-0.36 = log ([Na₂Suc] / [NaHSuc])

0.4365 = ([Na₂Suc] / [NaHSuc]) <em>(1)</em>

<em />

As total concentration of the buffer is 100mM = 0.100M:

0.100M = [Na₂Suc] + [NaHSuc] <em>(2)</em>

Replacing (2) in (1):

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 [NaHSuc] = 0.100M - [NaHSuc]

1.4365 [NaHSuc] = 0.100M

[NaHSuc] = 0.0696M

And:

[Na₂Suc] = 0.0304M

As volume of the buffer is 1L:

[NaHSuc] = 0.0696 moles

[Na₂Suc] = 0.0304 moles

Using molar mass of both substances:

Mass of monosodium succinate:

0.0696moles * (140g / 1mol) =<em> 9.744g of monosodium succinate.</em>

Mass of disodium succinate:

0.0304moles * (162g / 1mol) =<em> 4.925g of disodium succinate.</em>

<em></em>

5 0
2 years ago
Each orbit surrounding an atom is allowed _____.
Juli2301 [7.4K]
Two electrons is your answer glad to help!
4 0
2 years ago
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