All categories

3 years ago

Can anyone help with thiss??

Chemistry

1 answer:

IrinaVladis [17]3 years ago

5 0

l
kg
cm
gm
ml
km,m
m
nm
mg
ms
gm
nm
.
.
l,gm
.
kg
m
gm,m
s

Hope this helps :-). And....

ml: milliliter

kg: kilogram

m: meter

mg: milligram

mm: millimeter

ms: millisecond

l: liter

km: kilometer

µg: micro gram

cm: centimeter

g: gram

nm: nano meter

You might be interested in

What happens to water when it warms above 4 C

xxMikexx [17]

At 4 °C, the clusters start forming. The molecules are still slowing down and coming closer together, but the formation of clusters makes the molecules be further apart. Thus, the density of water is a maximum at 4 °C.

8 0

3 years ago

A tank with volume of 2.4 cu ft is filled with Methane to a pressure of 1500 psia at 104 degrees F. Determine the molecular weig

Soloha48 [4]

Explanation:

It is known that equation for ideal gas is as follows.

PV = nRT

The given data is as follows.

Pressure, P = 1500 psia, Temperature, T = $104^{o}F$ = 104 + 460 = 564 R

Volume, V = 2.4 cubic ft, R = 10.73 $psia ft^{3}/lb mol R$

Also, we know that number of moles is equal to mass divided by molar mass of the gas.

n = $\frac{mass}{\text{molar mass}}$

m = $n \times W$

= $0.594 \times 16.04$

= 9.54 lb

Hence, molecular weight of the gas is 9.54 lb.

We will calculate the density as follows.

d = $\frac{PM}{RT}$

= $\frac{1500 \times 16.04}{10.73 \times 564}$

= 3.975 $lb/ft^{3}$

Now, calculate the specific gravity of the gas as follows.

Specific gravity relative to air = $\frac{\text{density of methane}}{\text{density of air}}$

= $\frac{3.975 lb/ft^{3}}{0.0765 lb/ft^{3}}$

= 51.96

6 0

3 years ago

Nitrogen forms a surprising number of compounds with oxygen. A number of these, often given the collective symbol NOx (for "nitr

kvv77 [185]

Answer:

9.2

Explanation:

Let's do an equilibrium chart of this reaction:

2NO(g) + O₂(g) ⇄ 2NO₂(g)

4.9 atm 5.1 atm 0 Initial

-2x -x +2x Reacts (stoichiometry is 2:1:2)

4.9-2x 5.1-x 2x Equilibrium

The mole fraction of NO₂ (y) can be calculated by the Raoult's law, that states that the mole fraction is the partial pressure divided by the total pressure:

y = 2x/(4.9 - 2x + 5.1 -x + 2x)

0.52 = 2x/(10 - x)

2x = 5.2 -0.52x

2.52x = 5.2

x = 2.06 atm

Thus, the partial pressure at equilibrium are:

pNO = 4.9 -2*2.06 = 0.78 atm

pO₂ = 5.1 - 2.06 = 3.04 atm

pNO₂ = 2*2.06 = 4.12 atm

Thus, the pressure equilibrium constant Kp is:

Kp = [(pNO₂)²]/[(pNO)²*(pO₂)]

Kp = [(4.12)²]/[(0.78)²*3.04]

Kp = [16.9744]/[1.849536]

Kp = 9.2

4 0

3 years ago

You place 12.0 milliliters of water in a graduated cylinder. Then you add 15.0 grams of metal to the water and the new water lev

Marrrta [24]

Answer:

<h3>The answer is 7.85 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

volume = final volume of water - initial volume of water

volume = 13.91 - 12 = 1.91 mL

We have

$density = \frac{15}{1.91} \\ = 7.853403141...$

We have the final answer as

Hope this helps you

8 0

3 years ago

Read 2 more answers

9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8

MissTica

Answer: 2.24 grams of Pb

Explanation:

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles * 1 ) 1

Moles of Pb =0.0108 moles

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

5 0

2 years ago