All categories

4 years ago

Why are unused chemicals never returned to thier original containers?

Chemistry

1 answer:

NISA [10]4 years ago

3 0

Because if they were, they could contaminate the chemicals in the original container.

You might be interested in

A _______ reaction involves burning. this reaction requires energy to start, but it releases more energy then is required to sta

GenaCL600 [577]

The answer is combustion

5 0

3 years ago

Read 2 more answers

A research balloon at ground level contains 12 L of helium (He) at a pressure of 725mmHg and a temperature of 30.00∘C. When the

Helga [31]

<u>Answer:</u> The temperature when the volume and pressure has changed is -27.26°C

<u>Explanation:</u>

To calculate the temperature when pressure and volume has changed, we use the equation given by combined gas law. The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=725mmHg\\V_1=12L\\T_1=30.00^oC=[30+273]K=303K\\P_2=252mmHg\\V_2=28L\\T_2=?K$

Putting values in above equation, we get:

$\frac{725mmHg\times 12L}{303K}=\frac{252mmHg\times 28L}{T_2}\\\\T_2=\frac{252\times 28\times 303}{725\times 12}=245.74K$

Converting this into degree Celsius, we get:

$T(K)=T(^oC)+273$

$245.74=T(^oC)+273\\T(^oC)=-27.26^oC$

Hence, the temperature when the volume and pressure has changed is -27.26°C

5 0

3 years ago

What is the density of CHCL3 vapor at 1.00atm and 298K?

Advocard [28]

Answer:

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.

Explanation:

By ideal gas equation:

$PV=nRT$

Number of moles (n)

can be written as: $n=\frac{m}{M}$

where, m = given mass

M = molar mass

$PV=\frac{m}{M}RT\\\\PM=\frac{m}{V}RT$

where,

$\frac{m}{V}=d$ which is known as density of the gas

The relation becomes:

$PM=dRT$ .....(1)

We are given:

M = molar mass of chloroform= 119.5 g/mol

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the gas = $298K$

P = pressure of the gas = 1.00 atm

Putting values in equation 1, we get:

$1.00atm\times 119.5g/mol=d\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\d=4.88g/L$

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.

7 0

3 years ago

PLEASE ANSWER + BRAINLIEST !!!

Sunny_sXe [5.5K]

Hope this would help you

3 0

3 years ago

Anybody wana do this? Lol much appreciated.

Alex73 [517]

No units are provided in the data table, though one would usually assume that 0.244 is molarity of NaOH, and buret data of 14.7 is probably the final volume of NaOH in mL. No identification is given for the 38, but maybe this is the initial buret volume in mL.

If the buret was rinsed with water, it should have been left to dry for some time, or else the added water may slightly dilute the NaOH solution placed in it.

6 0

3 years ago

Read 2 more answers