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3 years ago

What kind of thing can be measured in derived units?

Chemistry

1 answer:

sergejj [24]3 years ago

7 0

Answer:

In this SI units system, there are seven SI base units and three supplementary units. The base SI units are metre, kilogram, second, kelvin, ampere, candela and the mole and the three supplementary SI units are radian, steradian and becquerel. All other SI units can be derived from these base units.

Explanation:

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mihalych1998 [28]

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4 0

2 years ago

What monatomic ions do selenium (z = 34) and phosphorus (z = 15) form? (type your answer using the format [cu]+ for cu+ and [cu]

Alex787 [66]

The answer:
for the monoatomic selenium ions
 -the ion charge of selenium is 2-, so the answer is [Se]2+
as for the monoatomic phosphorus ions
-the ion charge of phosphorus is 3-, so the answer is [P]3-

6 0

2 years ago

A gaseous compound is 30.4 % N and 69.6% OF. A 5.25 g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 958

Harman [31]

Answer:

The molecular formula = N2O4

Explanation:

Step 1: Data given

A gaseous compound is 30.4 % N and 69.6%

Mass of the compound = 5.25 grams

Volume of the gas = 1.00 L

Pressure of the gas = 958 mmHg = 1.26 atm

Temperature of the gas = -4 °C = 273 -4°C = 269 Kelvin

Molar mass of N = 14 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate mass of N

Mass of Nitrogen = 5.25 grams * 0.304 = 1.596 grams

Step 3: Calculate mass of O

Mass of Oxygen = 5.25 grams * 0.696 = 3.654 grams

Step 4: Calculate number of moles N

Number of moles N = Mass of N/ Molar mass of N

Moles of N = 1.596 grams / 14g/mol

Moles of N = 0.114 moles N

Step 5: Calculate moles of O

Moles O = 3.654 grams / 16 g/mol

Moles 0 = 0.2884 moles

Step 6: Calculate empirical formule

We calculate the empirical formule by dividing number of moles by the smallest number of mol

N : 0.114 / 0.114 = 1

O: 0.2284 / 0.114 = 2

Empirical formule = NO2

Step 7: Calculate number of moles of 5.25 g sample via gas law:

p*V = nRT

⇒ with p = the pressure = 1.26 atm

⇒ with v = 1.00 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/ K*mol

⇒ with T = the temperature = 269 K

number of moles n = (p*V)/(R*T)

n = (1.26*1L)/(0.08206*269)

n = 0.057 mol

Step 8: Calculate molar mass of the compound

This means 5.25 grams of the gas = 0.057 moles

So 1 mol of the compound has a molar mass of: 5.25 / 0.057 = 92.11 g/mol

Step 9: Calculate molar mass of the empirical formula NO2

N = 14 g/mol

O = 16 g/mol

NO2 = 14 + 16 + 16 = 46 g/mol

The empirical formule NO2 has a molar mass of 46 g/mol

Step 10: Calculate molecular formula

92.11 / 46 = 2

This means the empirical formula should be multiplied by 2

2*(NO2) = N2O4

The molecular formula = N2O4

8 0

2 years ago

The mole fraction of a non-electrolyte (MM 40.0 g/mol) in a saturated aqueous solution is 0.310. What is the molality of the sol

jeka57 [31]

Answer: The molality of non-electrolyte is 24.69 m

Explanation:

We are given:

Mole fraction of saturated aqueous solution = 0.310

This means that 0.310 moles of non-electrolyte is present.

Moles of water (solvent) = 1 - 0.310 = 0.690 moles

To calculate the mass from given number of moles, we use the equation:

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

Moles of water = 0.690 moles

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

$0.690mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.690mol\times 18g/mol)=12.42g$

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{n_{solute}\times 1000}{W_{solvent}\text{ (in grams)}}$

Where,

$n_{solute}$ = Moles of solute (non-electrolyte) = 0.310 moles

$W_{solvent}$ = Mass of solvent (water) = 12.42 g

Putting values in above equation, we get:

$\text{Molality of non-electrolyte}=\frac{0.310\times 1000}{12.42}\\\\\text{Molality of non-electrolyte}=24.96m$

Hence, the molality of non-electrolyte is 24.69 m

4 0

2 years ago

Find the ph of of 100 ml of an aqueous 0.43m baoh2 solution

denpristay [2]

Answer is: pH of barium hydroxide is 13.935.
Chemical dissociation of barium hydroxide in water:
Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻(aq).
c(Ba(OH)₂) = 0.43 M.
V(Ba(OH)₂) = 100 mL ÷ 1000 mL/L = 0.1 L.
n(Ba(OH)₂) = 0.43 mol/L · 0.1 L.
n(Ba(OH)₂) = 0.043 mol.
From chemical reaction: n(Ba(OH)₂) : n(OH⁻) = 1 : 2.
n(OH⁻) = 0.086 mol.
c(OH⁻) = 0.86 mol/L.
pOH = -logc(OH⁻).
pOH = 0.065.
pH = 14 - 0.065 = 13.935.

4 0

2 years ago