Which group has eight valence electrons?

If I have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 Liters, what is the temperature in Kelvin(K)

kvv77 [185]

Answer:

205 K (to 3 significant figures)

Explanation:

Assuming that 4 moles of the gas behaves like an ideal gas and obey the kinetic molecular theory.

Let's apply the ideal gas law, pV= nRT.

Here p denotes the pressure of the gas, V is for volume, n is the number of moles of the gas, R is the universal gas constant and T is the temperature.

Substitute the given information into the equation:

5.6 atm ×12 L= 4 mol ×R ×T

Since pressure is in atm and volume is in L, we can use R= 0.08206 L atm K⁻¹ mol⁻¹.

5.6 atm ×12 L= 4 mol ×0.08206 L atm K⁻¹ mol⁻¹ ×T

T= 67.2 ÷0.32824

T= 204.73 (5 s.f.)

T= 205 K (3 s.f.)

4 0

2 years ago

How many moles of water were lost if the amount of water lost was 0.456 grams? Do not include units and assume three significant

nika2105 [10]

<h3>Answer:</h3>

0.0253 mol H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 0.456 g H₂O (water)

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 0.456 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 0.025305 \ mol \ H_2O$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.025305 mol H₂O ≈ 0.0253 mol H₂O

3 0

2 years ago

Does 1 gram of phosphorus react with 6 grams of iodine to form 4 grams of phosphorus triodine in P4(s)+6I2(s)=4PI3(s)

mafiozo [28]

Answer:

No

Explanation:

One mole of P₄ react with six moles of I₂ and gives 4 moles of PI₃.

When one gram phosphorus and 6 gram of iodine react they gives 8.234 g ram of PI₃ .

Given data:

Mass of phosphorus = 1 g

Mass of iodine = 6 g

Mass of PI₃ = ?

Solution:

Chemical equation:

P₄ + 6I₂ → 4PI₃

Number of moles of P₄:

Number of moles = Mass /molar mass

Number of mole = 1 g / 123.9 g/mol

Number of moles = 0.01 mol

Number of moles of I₂:

Number of moles = Mass /molar mass

Number of moles = 6 g / 253.8 g/mol

Number of moles = 0.024 mol

Now we will compare the moles of PI₃ with I₂ and P₄.

I₂ : PI₃

6 : 4

0.024 : 4/6×0.024 = 0.02

P₄ : PI₃

1 : 4

0.01 : 4 × 0.01 = 0.04 mol

The number of moles of PI₃ produced by I₂ are less it will be limiting reactant.

Mass of PI₃ = moles × molar mass

Mass of PI₃ = 0.02 mol × 411.7 g/mol

Mass of PI₃ = 8.234 g

4 0

2 years ago

Calculate the amount of oxygen gas collected by the displacement of water at 14◦C if the atmospheric pressure is 790 Torr and th

zhannawk [14.2K]

Answer : The amount of oxygen gas collected are, 0.217 mol

Explanation :

Using ideal gas equation :

$PV=nRT$

where,

P = pressure of gas = $(790-12)torr=778torr=1.02atm$ (1 atm = 760 torr)

V = volume of gas = 5 L

T = temperature of gas = $14^oC=273+14=287K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1.02atm)\times (5L)=n\times (0.0821L.atm/mol.K)\times (287K)$

$n=0.217mole$

Thus, the amount of oxygen gas collected are, 0.217 mol

8 0

3 years ago

Which need, want value, or interest was most likely involved in the

____ [38]

For this question the answer is c

8 0

2 years ago

Read 2 more answers