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Tems11 [23]
2 years ago
12

Can someone put their own numbers in this sentence and solve it? Thanks!

Chemistry
1 answer:
Nat2105 [25]2 years ago
4 0

A balloon with a pressure of 1 atm and a temperature of 17° Celsius has a volume of 1000 mL. The day warms up and increases the temperature to 27° Celsius, which makes the pressure change to 1520 mm Hg.What is the new volume of the balloon, in Liters?

The new volume of the balloon is <u>0.517 L.</u>

By ideal gas law, PV = nRT

P = pressure

V = volume

n = number of mole of gas

R = gas constant

T = temperature

Using ideal gas law, P₁V₁/ T₁ = P₂V₂/ T₂

P₁= 1 atm

V₁ = 1000 mL = 1 L

T₁ = 17° C = 290 K

After the day warms up:-

P₂ = 1520 mm Hg = 2 atm

T₂ = 27° C = 300 K

Put these values in formula, P₁V₁/ T₁ = P₂V₂/ T₂

V₂ =  P₁V₁T₂/ T₁ P₂

V₂ = 1 atm× 1 L × 300 K/ 290 K × 2 atm

V₂ = 0.517 L = 517 mL

Hence, the new volume of the balloon is 0.517 L.

The ideal gas equation is formulated as: PV = nRT.

In this equation, P refers to the pressure of the ideal gas, V is the volume of the ideal gas, n is the overall quantity of ideal gas this is measured in terms of moles, R is the universal gas constant, and T is the temperature.

Learn more about ideal gas law here:- brainly.com/question/25290815

#SPJ1

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The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge,
Mila [183]

Answer:

1.33 Å

Explanation:

Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å

Also,

For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.

Thus,

\frac {r^+}{r^-}=0.731 .................1

Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.

Thus,

r^++r^-=\frac {a}{2}  ...................2

Given that:

Cl^-\ (r^-) = 1.82\ \dot{A}

To find,

K^+\ (r^+) = ? \dot{A}

Using 1 and 2 , we get:

1.731\ r^+=0.731\times \frac {6.28}{2}

<u>Size of the potassium ion = 1.33 Å</u>

4 0
3 years ago
In terms of their electron configurations, why is cesium more likely to lose its valence electron than potassium?
Harlamova29_29 [7]

Explanation:

use the term electron sheilding, the more electrons between the valence el3ctron and nucleus the easier to lose the valence electron (more sheilding = easier to lose)

5 0
3 years ago
A sample of a compound contains ​0.300 mol​ ​of carbon​ and ​1.20 mol​ ​of hydrogen​. What is the ​empirical formula of the comp
Margarita [4]

Answer:

CH4

Explanation:

The number of moles of carbon and hydrogen has been given as follows:

C = ​0.300 mol​ ​

H = ​1.20 mol

Next, we divide each mole value by the smallest (0.300)

C = 0.300 ÷ 0.300 = 1

H = 1.20 ÷ 0.300 = 4

The empirical ratio of Carbon and Hydrogen is 1:4, hence, the empirical formula is CH4

3 0
3 years ago
If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.
vesna_86 [32]

Answer:

Sr(OH)2

Explanation:

We'll begin by calculating the number of mole of carbonic acid in 150mL of 3.5 M carbonic acid solution. This is illustrated below:

Molarity = 3.5M

Volume = 150mL = 150/1000 = 0.15L

Mole of carbonic acid, H2CO3 =..?

Mole = Molarity x Volume

Mole of carbonic acid, H2CO3 = 3.5 x 0.15 = 0.525 mole.

Next, we shall convert 0.525 mole of carbonic acid, H2CO3 to grams.

Mole of H2CO3 = 0.525 mole

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 =..?

Mass = mole x molar mass

Mass of H2CO3 = 0.525 x 62 = 32.55g

Next, we shall write the balanced equation for the reaction. This is given below:

Sr(OH)2 + H2CO3 → SrCO3 + 2H2O

Next, we shall determine the mass of Sr(OH)2 and H2CO3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Sr(OH)2 = 88 + 2(16 + 1) = 88 + 2(17) = 122g/mol

Mass of Sr(OH)2 from the balanced equation = 1 x 122 = 122g

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 from the balanced equation = 1 x 62 = 62g.

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Therefore, 12.5g of Sr(OH)2 will react with = (12.5 x 62)/122 = 6.35g.

We can see evidently from the calculations made above that it will take 6.35g out 32.55g of H2CO3 to react with 12.5g of Sr(OH)2. Therefore, Sr(OH)2 is the limiting reactant and H2CO3 is the excess reactant

5 0
3 years ago
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dedylja [7]

Answer:

B

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its an acidic oxide, it disolves in water to form carbonic acid which is an acid

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