All categories

4 years ago

A gas occupies a volume of 4.50 L at 44 kPa. What would be the new volume at 50 kPa?

Chemistry

1 answer:

oksano4ka [1.4K]4 years ago

5 0

Explanation:

P1= 44 kpa

P2= 50 kpa

V1= 4.50 L

V2= ?

P1 V1= P2 V2

44 × 4.50 = 50 × V2

198= 50 × V2

V2 = 198/ 50

V2= 3.96 L "the new volume"

You might be interested in

Describe how to draw a 3-heptyne

lesantik [10]

You can't really describe it but this is what it looks like http://www.chemspider.com/Chemical-Structure.453291.html

3 0

3 years ago

All the ways scientists do investigations

Eva8 [605]

Scientific metod this is one of them

3 0

3 years ago

Calculate the mass of one mole of these substances.

pogonyaev

Answer:

a. 53.5 g/mol

b. 80.06 g/mol

c. 133.33 g/mol

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Molar Mass - 1 mol per x grams substance

Explanation:

Step 1: Define

a. NH₄Cl

b. NH₄NO₃

c. AlCl₃

Step 2: Find masses

Molar Mass of N - 14.01 g/mol

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Al - 26.98 g/mol

Molar Mass of Cl - 35.45 g/mol

Step 3: Calculate compound masses

Molar Mass of NH₄Cl - 14.01 g/mol + 4(1.01 g/mol) + 35.45 g/mol = 53.5 g/mol

Molar Mass of NH₄NO₃ - 2(14.01 g/mol) + 4(1.01 g/mol) + 3(16.00 g/mol) = 80.06 g/mol

Molar Mass of AlCl₃ - 26.98 g/mol + 3(35.45 g/mol) = 133.33 g/mol

6 0

3 years ago

On the space shuttle, lithium hydroxide is produced from lithium oxide and water. The lithium hydroxide then reacts with the exh

Ratling [72]

Answer:

6,2g of CO₂

Explanation:

Based on the reactions:

Li₂O(s) + H₂O(g) → 2LiOH(s)

LiOH(s) + CO₂(g) → LiHCO₃(s)

2,6g of Li₂O and 1,3g of H₂O are:

2,6g × ( 1mol / 29,88g) = 0,087 moles

1.3g × ( 1mol / 18,01g) = 0,072 moles

That means limiting reactant is H₂O. The moles produced of LiOH are:

0,072 moles of H₂O × ( 2mol LiOH / 1mol H₂O) = 0,14 moles of LiOH

Thus, the maximum CO₂ that can react are 0,14 moles of CO₂, in grams

0,14 moles CO₂ × (44,01g / 1mol) = 6,2g of CO₂

I hope it helps!

6 0

3 years ago

A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita

mestny [16]

Answer:

The answer is:

(a) $NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)$

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of $AgNO_3$ ,

= 30.0 ml

or,

= 0.030 L

Molarity of $AgNO_3$ ,

= 0.050 M

Moles of NaCl will be:

= $\frac{Given \ mass}{Molar \ mass}$

= $\frac{0.0860}{58.44}$

= $0.00147 \ mol$

now,

Moles of $AgNO_3$ will be:

$= Molarity\times Volume$

$= 0.050\times 0.030$

$=0.0015 \ mol$

(a)

The reaction is:

⇒ $NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)$

(b)

1 mole of NaCl react with,

= 1 mol of $AgNO_3$

0.0015 mol $AgNO_3$ needs,

= $0.00150 \ mol \ NaCl$

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= $0.00147\times \frac{1}{1}\times \frac{143.32}{1}$

= $0.211 \ g \ AgCl$

4 0

3 years ago