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Aleks [24]
3 years ago
10

I need help, idk how to do PV=nRT

Chemistry
1 answer:
Likurg_2 [28]3 years ago
6 0

Answer:

107.64L

Explanation:

V= Volume in Liter

T= Temp in Kelvin (degrees celcius + 275)

N= # Mols

P= Pressure

R= Gas Constant for atm is .0821

so

for P

V=NRT/P

(4x,0821x295)/.900=107.64L

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look in your science book

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2 years ago
What is the bohr model for pottasium
guapka [62]
Potassium is an alkali metal with the chemical symbol K. It has an atomic number of 19, meaning that it has 19 positively charged protons. It also contains 19 electrons, which have a negative charge, and 20 neutrons, which do not hold a charge
Hope this help
7 0
3 years ago
Which of the following is an organic moleclue ?
Doss [256]

Answer:

water

Explanation:

6 0
3 years ago
The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K,
ludmilkaskok [199]

Answer: 4.3\times 10^{-13}s^{-1}

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 775.0 = ?

Ea = activation energy for the reaction = 262 kJ/mol = 262000J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 775.0K

Now put all the given values in this formula, we get

\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]

\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150

(\frac{6.1\times 10^{-8}}{K_2})=141253.8

Therefore, the value of the rate constant at 775.0 K is 4.3\times 10^{-13}s^{-1}

5 0
3 years ago
A sample of a compound is determined to have 1.17 g of carbon and 0.287 g of hydrogen. what is the correct representation of the
yarga [219]

CH3 is the empirical formula for the compound.

A sample of a compound is determined to have 1.17g of Carbon and 0.287 g of hydrogen.

The number of atom or moles in the compound is

1.17 g C X  1 mol of C / 12.011 g C = 0.097411 mol of C.

0.287 g H x 1 mol of  H / 1 g H = 0.28474 mol H.

This compound contains 0.097411 mol of carbon and 0.28474 mol of Hydrogen.

So we can represent the compound with the formula C0.974H0.284.

Subscripts in formulas can be made into whole numbers by multiplying the smaller subscript by the larger subscript.

we can divide 0.284 by 0.0974.

0.284 / 0.0974 = 3.

So here, Carbon is one and hydrogen is 3.

We can write the above formula as a CH3.

Hence the empirical formula for the sample compound is CH3.

For a detailed study of the empirical formula refer given link brainly.com/question/13058832.

#SPJ1.

5 0
2 years ago
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