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Bas_tet [7]
3 years ago
7

Identify the molar mass of each of the following.

Chemistry
1 answer:
Zepler [3.9K]3 years ago
4 0
Cr: 51.98610 g/mol
CH4: 15.0425 g/mol
AI2(SO4)3: 342.1509 g/mol
Br2: 159.8080 g/mol
You might be interested in
A gas mixture contains 1.52 atm of Ne, 766 mmHg of He and Ar. What is the partial pressure, in atmospheres, of At if the gas mix
ValentinkaMS [17]

Answer:

0.74 atm.

Explanation:

From the question given above, the following data were obtained:

Pressure of Ne (Pₙₑ) = 1.52 atm

Pressure of He (Pₕₑ) = 766 mmHg

Total pressure (Pₜ) = 3.27 atm

Pressure of Ar (Pₐᵣ) =?

Next, we shall convert the pressure of He from mmHg to atm. This can be obtained as follow:

760 mmHg = 1 atm

Therefore,

766 mmHg = 766 mmHg × 1 atm / 760 mmHg

766 mmHg = 1.01 atm

Finally, we shall determine the partial pressure of Ar. This can be obtained as follow:

Pressure of Ne (Pₙₑ) = 1.52 atm

Pressure of He (Pₕₑ) = 1.01 atm

Total pressure (Pₜ) = 3.27 atm

Pressure of Ar (Pₐᵣ) =?

Pₜ = Pₙₑ + Pₕₑ + Pₐᵣ

3.27 = 1.52 + 1.01 + Pₐᵣ

3.27 = 2.53 + Pₐᵣ

Collect like terms

3.27 – 2.53 = Pₐᵣ

Pₐᵣ = 0.74 atm

Thus the partial pressure of Ar is 0.74 atm.

8 0
3 years ago
What's the "p" process in chemistry
Flauer [41]
The proton capture process
6 0
3 years ago
The acidic substance in vinegar is acetic acid, HC2H3O2. When 4.00 g of a certain vinegar sample was titrated with 0.200 M NaOH,
mihalych1998 [28]

Answer:

7.7%

Explanation:

the balanced equation for the neutralization reaction is:

CH3COOH+NaOH→CH3COONa+H2O

from the balanced equation above,the mole ratio of CH3COOH to NaOH is 1:1

no of mole of CH3COOH= no of mole of NaOH

no of mole of NaOH= concentration(M) × Volume(L)

=0.2 M× 0.02556=0.005112 mol

since, no of mole of CH3COOH= no of mole of NaOH

no of mole of CH3COOH=0.005112 mol

However ,no of mole =mass/molar mass

molar mass of CH3COOH=60.052g/mol

mass of CH3COOH=0.005112×60.052

=0.307g

percentage by mass of acetic acid in the vinegar=0.307/4

=0.07675×100%

=7.675%

5 0
4 years ago
A student is investigating acceleration using balls with
netineya [11]

Answer:

15

Explanation:

3 0
3 years ago
Rolls of foil are 304 mmmm wide and 0.014 mmmm thick. (The density of foil is 2.7 g/cm3g/cm3 .) What maximum length of foil can
ratelena [41]

Answer:

92.2 m

Explanation:

Given that:=

The breadth = 304 mm

Height = 0.014 mm

Let Length = x mm

Volume = Length\times breadth\times height

Thus,

Volume = 304\times 0.014\times x\ mm^3=4.256x\ mm^3

Also, 1 mm³ = 0.001 cm³

So, volume = 0.004256 cm³

Given that density = 2.7 g/cm³

Mass = 1.06 kg = 1060 g

So,

Volume=\frac{Mass}{Density}=\frac{1060}{2.7}\ cm^3=392.59\ cm^3

So,

0.004256*x = 392.59

x = 92243.89 mm

Length of foil = 92243.89 mm = 92.2 m

3 0
3 years ago
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