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oee [108]
3 years ago
7

The ideal gas constant, R has several different values that could be used. Which quantity causes these differences?

Chemistry
1 answer:
givi [52]3 years ago
4 0

Answer:

\boxed{\text{Pressure and volume}}

Explanation:

The measure of R is always the same, but the numbers may differ depending on the units you use.

For example, in SI units, R = 8.314 Pa·m³K⁻¹mol

If your measurement uses different units, you must either convert your units to SI or use a value of R consistent with your units.

If you use bars and litres,                       R = 0.083 14   bar·L·K⁻¹mol⁻¹.

If you use kilopascals and litres,            R = 8.314        kPa·L·K⁻¹mol

If you use atmospheres and litres,        R = 0.082 06 L·atm·K⁻¹mol⁻¹.

If you use Torr and cubic centimetres, R = 62 368     Torr·cm³ K⁻¹mol⁻¹.

The only units that don't change are "K⁻¹mol⁻¹".

\text{The quantities that affect the value of R are }\boxed{\textbf{pressure and volume}}

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Answer:

15

Explanation:

8 0
3 years ago
Which of the Atoms shown has an atomic number four
dimulka [17.4K]

Answer:

B

Explanation:

Atomic # = Protons

it says 4 p in the inside of the orbital

4 0
3 years ago
1. NaOH mass of a solution of 200g in which its percentage is 25%. What mass of sulfuric acid solution is needed to completely n
Ede4ka [16]

Answer:

m_{H2SO4 = 61.25 g

m_{Na2SO4} = 88.75 g

Explanation:

m_{NaOH} = \frac{200 . 25 }{100} = 50 g

⇒ n_{NaOH} = \frac{50}{40} = 1.25 (moles)

2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O

   2        :     1           :      1         :    2

 1.25                                                       (moles)

⇒  n_{H2SO4} = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ m_{H2SO4} = 0.625 × 98 = 61.25 g

    n_{Na2SO4} = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒m_{Na2SO4} = 0.625 × 142 = 88.75 g

4 0
3 years ago
In a constant‑pressure calorimeter, 60.0 mL of 0.300 M Ba ( OH ) 2 was added to 60.0 mL of 0.600 M HCl . The reaction caused the
densk [106]

Answer:

Q sln = 75.165 J

Explanation:

a constant pressure calorimeter:

  • Q sln = mCΔT

∴ m sln = m Ba(OH)2 + m HCl

∴ molar mass Ba(OH)2 = 171.34 g/mol

∴ mol Ba(OH)2 = (0.06 L)(0.3 mol/L) = 0.018 mol

⇒ mass Ba(OH)2 = (0.018 mol)(171.34 g/mol) = 3.084 g

∴ molar mass HCl = 36.46 g/mol

∴ mol HCl = (0.06 L)(0.60 mol/L) = 0.036 mol

⇒ mass HCl = (0.036 mol)(36.46 g/mol) = 1.313 g

⇒ m sln = 3.084 g + 1.313 g = 4.3966 g

specific heat (C):

∴ C sln = C H2O = 4.18 J/g°C

∴ ΔT = 26.83°C - 22.74°C = 4.09°C

heat absorbed (Q):

⇒ Q sln = (4.3966 g)(4.18 J/g°C)(4.09°C)

⇒ Q sln = 75.165 J

8 0
3 years ago
Calculate the missing variables in each experiment below using Avogadro’s law.
blagie [28]

Answer:

The answer to your question is: letter c

Explanation:

Data

V1 = 612 ml    n1 = 9.11 mol

V2 = 123 ml    n2 = ?

Formula

                               \frac{V1}{n1}  =  \frac{V2}{n2}

                                         n2 = \frac{n1V2}{V1}

                                         n2 = \frac{(9.11)((123)}{(612)}

                                                n2 = 1.83 mol                                                

5 0
3 years ago
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