As I am reading the problem,i can tell the question gives you two temperatures, two pressures, one volume and asking for the other. this should be an indication that you need to use the following gas formula

P1V1/T1= P2V2/T2

P1= 2.0 atm
V2= 4.0 L
T1= 27= 300 K
P2= 8.0 atm
V2= ?
T2= 327= 600 k

let's plug in the values into the formula

(2.0 x 4.0)/ 300= (8.0 x V2)/ 600 K

V2= 2.0 Liters

answer is D

3 0

3 years ago

Question: Laundry in a clothing dryer often become charged with static electricity while drying. Which of these best explains wh

zavuch27 [327]

The answer to this question will be D

7 0

3 years ago

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If 175 g of phosphoric acid reacts with 150.0 g of sodium hydroxide, what is the limiting reactant? How many grams of sodium pho

Evgen [1.6K]

Answer:

NaOH is the limiting reactant.

204.9 g of sodium phosphate are formed.

51.94 g of excess reactant will remain.

Explanation:

The reaction that takes place is:

H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O

First we convert the mass of both reactants to moles, using their respective molar masses:

H₃PO₄ ⇒ 175 g ÷ 98 g/mol = 1.78 mol

NaOH ⇒ 150 g ÷ 40 g/mol = 3.75 mol

1.78 moles of H₃PO₄ would react completely with (1.78 * 3) 5.34 moles of NaOH. There are not as many NaOH moles so NaOH is the limiting reactant.

We calculate the produced moles of Na₃PO₄ using the limiting reactant:

3.75 mol NaOH * $\frac{1molNa_3PO_4}{3molNaOH}$ = 1.25 mol Na₃PO₄

Then we convert moles into grams:

1.25 mol Na₃PO₄ * 163.94 g/mol = 204.9 g

We calculate how many H₃PO₄ moles would react with 3.75 NaOH moles:

3.75 mol NaOH * $\frac{1molH_3PO_4}{3molNaOH}$ = 1.25 mol H₃PO₄

We substract that amount from the original amount:

1.78 - 1.25 = 0.53 mol H₃PO₄

Finally we convert those remaining moles to grams:

0.53 mol H₃PO₄ * 98 g/mol = 51.94 g

3 0

3 years ago