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4 years ago

13

A diatomic gas has the molar mass 38.00 g/mol. What is a possible formula?

1 answer:

Alinara [238K]4 years ago

7 0

It’s Florine! And the formula is F subscript 2

You might be interested in

How many grams are in 5.2 moles of Li2SO4

skelet666 [1.2K]

Answer:

572 g

Explanation:

Molar mass is the mass of 1 mol of an element or compound

molar mass of Li₂SO₄ is the sum of the products of the molar masses of the elements by the number of atoms in the compound

molar masses of each element making up lithium sulphate

Li - 7 g/mol

S - 32 g/mol

O - 16 g/mol

molar mass of Li₂SO₄ - (7 g/mol x 2) + ( 32 g/mol x 1) + ( 16 g/mol x 4 )

molar mass = 110 g/mol

mass of 1 mol of Li₂SO₄ is 110 g

therefore mass of 5.2 mol of Li₂SO₄ is - 110 g/mol x 5.2 mol = 572 g

mass is 572 g

7 0

3 years ago

if the density of a bar of gold is 19.3g/cm^3and you cut it into four pieces , what is its density of each piece of gold

AlladinOne [14]

Density is an intrinsic property, so it is independent of the amount of substance present: one gold coin would have the same density as a solid gold boulder.

So if the density of gold is 19.3 g/cm³, the density of a bar of gold and the pieces into which the bar is cut would all be 19.3 g/cm³.

4 0

4 years ago

Name the two products when calcium carbonate (CaCO3) is heated

trapecia [35]

Answer:

The products are Calcium oxide and Carbon dioxide.

Explanation:

When calcium carbonate is heated, thermal decomposition occurs.

Calcium calcium → Calcium oxide + Carbon dioxide

3 0

3 years ago

Read 2 more answers

Plssssssssssssssssssmejepleplele

Leona [35]

Answer:

Yes the two of the answer is True

3 0

3 years ago

Read 2 more answers

Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

garri49 [273]

Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

= 4.6 bar

and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

Now, we will calculate the vapor composition as follows.

$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

4 0

4 years ago