Which beaker are the molecules are moving faster, and which beaker is it hotter?

What is the concentration of the solution prepared by dissolving 2.35 g of KBr (M = 119 g/mol) in 250 mL of water?

dusya [7]

Answer:

The concentration of KBr is $C = 0.07899 \ mol L^{-1}$

Explanation:

From the question we are told that

The mass of KBr is $m_{KBr} = 2.35 \ g$

The molar mass of KBr is $M_{KBr} = 119 g/mol$

Volume of water is $V = 250 \ mL = 250 *10^{-3} = 0.250 \ L$

This implies that the volume of the solution is $V = 250 mL$

The number of moles of KBr is

$n = \frac{m_{KBr}}{M_{KBr}}$

Substituting values

$n = \frac{2.35}{119}$

$n = 0.01975 \ mol$

The concentration of KBr is mathematically represented as

$C = \frac{0.01975}{0.250}$

$C = 0.07899 \ mol L^{-1}$

3 0

3 years ago

What eventually happens to gas if its pressureis increased?

balu736 [363]

Answer:

the rate increases

Explanation:

they are closer to each other, they will collide with each other more frequently and more successful collision

4 0

2 years ago

Rust forming on a piece of iron is an example of which of the following? *

kvasek [131]

Rust is an iron oxide and formed by the reaction of iron and oxygen in the presence of moisture. So the answer would be C

8 0

2 years ago

Which overall chemical equation is obtained by combining these intermediate equations?

Juli2301 [7.4K]

The common substance among the product(s) of the first equation and among the reactant(s) in the second equation is H2O(g). We can eliminate that as an intermediate. The overall chemical equation will thus be:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l),

which is the first answer choice.

In essence, all you’re doing here is swapping water vapor for liquid water.

5 0

2 years ago

What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91

Vikentia [17]

Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

$1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L$

Temperature = $91^{o}C = (91 + 273) K = 364 K$

As molar mass of $O_2$ is 32 g/mol. Hence, the number of moles of $O_2$ are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol$

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

$PV = nRT\\P \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm$

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

4 0

2 years ago