What is a single<br> replacement rxn?

At its critical point, ammonia has a density of 0.235 g cm23. You have a special thickwalled glass tube that has a 10.0mm outs

salantis [7]

Answer:

$\large \boxed{\text{69.3 mg}}$

Explanation:

1. Volume of sealed tube

Assume the sealed tube is a right circular cylinder in which the cap and the base are also 4.20 mm thick.

Its outside dimensions are 155 mm long × 10.0 mm diameter.

Its inside dimensions are

h = 155 mm - 2 × 4.20 mm = 146.6 mm

r = 5.0 mm - 4.20 mm = 0.8 mm

V = πr²h = π(0.8)²× 146.6 mm³ = 294.8 mm³ = 0.2948 cm³

2. Calculate the mass of NH₃

$\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\0.235 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{0.2948 cm}^{3}}\\\\m & = & \text{0.0693 g}\\& = & \textbf{69.3 mg}\\\end{array}\\\text{You must seal $\large \boxed{\textbf{69.3 mg}}$ of ammonia in the tube.}$

4 0

2 years ago

If 2.60 g of NaBr are dissolved in enough water to make 160. mL of solution, what is the molar concentration of

navik [9.2K]

This problem has two parts; the first one asking for the concentration of NaBr given both its mass and volume and the second one asking for its volume given both mass and concentration. The answers turn out to be 0.158 M and 211 mL.

<h3>Molarity</h3>

In chemistry, the use of units of concentration depends on both the substances to analyze and their amounts. In such a way, for molarity, one needs the following relationship between the moles of solute and volume of solution:

$M=\frac{n}{V}$

Thus, for the first part of the problem we first calculate the moles in 2.60 g of NaBr via its molar mass:

$2.60g*\frac{1mol}{102.89g} =0.0253mol$

Next, we convert the 160. mL to L by dividing by 1000 in order to obtain 0.160 L to subsequently calculate the molarity:

$M=\frac{0.0253mol}{0.160L}=0.158M$

Next, since the moles remain the same and for the second part we are asked for the volume given the concentration, one can solve for the volume so as to obtain:

$V=\frac{n}{M} =\frac{0.158M}{0.120mol/L}\\ \\V=0.211L$

That in milliliters turns out to be:

$V=0.211L*\frac{1000mL}{1L}=211mL$

Learn more about molarity: brainly.com/question/10053901

6 0

1 year ago

How many grams of Cl are in 485g of CaCl2

Oksana_A [137]

329g Cl (You're welcome :D)

6 0

2 years ago

20 mL of a gas at 200K is heated until the new volume is 55 mL. What is the final temperature of the gas?

Lostsunrise [7]

Answer:

550 K

Explanation:

Given the following data;

Original Volume, V1 = 20mL

Original Temperature, T1 = 200K

New Temperature, V2 = 55mL

To find new temperature T2, we would use Charles' law.

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

$VT = K$