All categories

3 years ago

The melting point of unknown substance A is 15°C and its boiling point is 95°C. Sketch a

Chemistry

1 answer:

Igoryamba3 years ago

4 0

Answer:

Idk if this is right but i hope it helps... sorry if it's wrong

Explanation:

You might be interested in

Which best describes a compound such as sodium chloride

Elza [17]

It is a compound composed of sodium (Na) and chloride (Cl)

5 0

3 years ago

If all data points lie in a straight line on your standard curve except for one point. Should you ignore the point or shift the

Ivanshal [37]

Ignore the point, the majority of the points are along the trend line.

3 0

3 years ago

Read 2 more answers

AGAIN PLEASE ONLY ANSWER IF YOUR GOING TO HELP TY!

Vladimir [108]

Answer:

do you have any socials to add me so i ca answer /try answer all ur questions or if not jist asl the internet these question

Explanation:

gvggghg

5 0

3 years ago

What element has only one proton in its nucleus??

blagie [28]

Hydrogen is the only element in entire periodic table with only 1 proton in its nucleus.

3 0

3 years ago

Nickel carbonyl, Ni(CO)4 is one of the most toxic substances known. The present maximum allowable concentration in laboratory ai

vodomira [7]

<u>Answer:</u> The mass of $Ni(CO)_4$ allowable in the laboratory is 4599.5 grams

<u>Explanation:</u>

To calculate the volume of cuboid, we use the equation:

$V=l\times b\times h$

where,

V = volume of cuboid

l = length of cuboid = 14 ft

b = breadth of cuboid = 22 ft

h = height of cuboid = 9 ft

Putting values in above equation, we get:

$V=14\times 22\times 9=2772ft^3=78503.04L$ (Conversion factor: $1ft^3=28.32L$

To calculate the moles of gas, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 1.00 atm

V = Volume of the gas = 78503.04 L

T = Temperature of the gas = $23^oC=[23+273]K=296K$

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

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

$1.00atm\times 78503.04L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 296K\\\\n=\frac{1.00\times 78503.04}{0.0821\times 296}=3230.4mol$

Applying unitary method:

For every 109 moles of gas, the moles of $Ni(CO)_4$ present are 1 moles

So, for 3230.4 moles of gas, the moles of $Ni(CO)_4$ present will be = $\frac{1}{109}\times 3230.4=26.94mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of $Ni(CO)_4$ = 26.94 moles

Molar mass of $Ni(CO)_4$ = 170.73 g/mol

Putting values in above equation, we get:

$26.94mol=\frac{\text{Mass of }Ni(CO)_4}{170.73g/mol}\\\\\text{Mass of }Ni(CO)_4=(26.94mol\times 170.73g/mol)=4599.5g$

Hence, the mass of $Ni(CO)_4$ allowable in the laboratory is 4599.5 grams

3 0

3 years ago