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3 years ago

What is the atomic weight of Na?

Chemistry

1 answer:

Zigmanuir [339]3 years ago

6 0

Answer:

22.989769 u

Explanation:

The atomic weight of any atom can be found by multiplying the abundance of an isotope of an element by the atomic mass of the element and then adding the results together. This equation can be used with elements with two or more isotopes: Carbon-12: 0.9889 x 12.0000 = 11.8668. Carbon-13: 0.0111 x 13.0034 = 0.1443.

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find the percent yield,if 6.0g of Licl are actually produced and the theoretical yield is 35.4g of Licl.

drek231 [11]

Answer:

Percent yield = 17%

Explanation:

Given data:

Actual yield of lithium chloride = 6.0 g

Theoretical yield of lithium chloride = 35.4 g

Percent yield = ?

Solution:

Formula:

Percent yield = (actual yield / theoretical yield )× 100

Now we will put the values in formula.

Percent yield = (6.0 g/ 35.4 g)× 100

Percent yield = 0.17 × 100

Percent yield = 17%

8 0

3 years ago

What is the product of the reaction between arsenic and oxygen

Ivan

Answer:

Explanation:

As + O2 -------->as2o3

5 0

3 years ago

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The molar mass of an unknown gas was measured by an effusion experiment. It was found that it took 60 s for the gas to effuse, w

Marizza181 [45]

Answer:

The molar mass of the gas is 44 g/mol

Explanation:

It is possible to solve this problem using Graham's law that says: Rates of effusion are inversely dependent on the square of the mass of each gas. That is:

$\frac{r_1}{r_2} =\frac{\sqrt{M_2} }{\sqrt{M_1} }$

If rate of effusion of nitrogen is Xdistance / 48s and for the unknown gas is X distance / 60s and mass of nitrogen gas is 28g/mol (N₂):

$\frac{X/48s}{X/60s} =\frac{\sqrt{M_2} }{\sqrt{28g/mol} }$

6,61 = √M₂

44g/mol = M₂

<em>The molar mass of the gas is 44 g/mol</em>

I hope it helps!

4 0

3 years ago

Wat is the meaning for autotrophs

mario62 [17]

An organism that is able to form nutritional organic substances from simple inorganic substances such as carbon dioxide.

6 0

3 years ago

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Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50

devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C $_{V}$ You can work out C $_{V}$

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC $_{V}$ ∆T

Polyatomic gas: C $_{V}$ = 3R

∆U= nC $_{V}$ ∆T

∆U= 28g x C $_{V}$ x (350K - 58.3K)

∆U = 28C $_{V}$ x 291.7

∆U = 10967.6 x C $_{V}$

5 0

3 years ago