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3 years ago

How many moles are in 4.24kg of Na2CO3?

Chemistry

1 answer:

frosja888 [35]3 years ago

8 0

40.0 is the answer, sorry, I had some lag, I would've answered it like 5 minutes ago

Hope I helped

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What is one molar volume of the gas ammonia (NH3) at STP ?

tatyana61 [14]

D 42.2 L/mol I’s is it I got it correct

7 0

2 years ago

Read 2 more answers

an atom of which element has the strongest attraction for electons in a chemical bond?cholorine, phosphorus, carbon, sulfur

vredina [299]

Answer:

D) Sulphur

Explanation:

Sulphur has the strongest electromagnetivity here

7 0

3 years ago

Hemoglobin is the protein that transports oxygen in mammals. Different species of mammals have slightly different forms of hemog

Anna35 [415]

Answer;

= 64561.95 g/mole

Explanation;

mass of Fe in 100g = .346g

= .346 / 55.8452 moles

= 0.0061957 moles

These represent 4 moles of Fe in the molecule so moles of hemaglobin

= 0.0061957/4

= 0.0015489 moles

these are in 100 g so mass of 1 mole = 100 / 0.0015489

= 64561.95 g / mole

molar mass of hemoglobin = 64561.95 g/mole

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3 years ago

Why is acid added to water during eletrolysis

DedPeter [7]

Answer:

Pure water is a non conductor of electricity and dilute acids in their aqueous solutions form free ions, which conducts electricity. Thus when we need to electrolyse water, a dilute acid is added to increase its conductivity.

4 0

3 years ago

Titanium has an HCP crystal structure, a c/a ratio of 1.669, an atomic weight of 47.87 g/mol, and a density of 4.51 g/cm3. Compu

IrinaK [193]

Answer : The atomic radius for Ti is, $1.45\times 10^{-8}cm$

Explanation :

Atomic weight = 47.87 g/mole

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

First we have to calculate the volume of HCP crystal structure.

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times V}$ .............(1)

where,

$\rho$ = density = $4.51g/cm^3$

Z = number of atom in unit cell (for HCP = 6)

M = atomic mass = 47.87 g/mole

$(N_{A})$ = Avogadro's number

V = volume of HCP crystal structure = ?

Now put all the values in above formula (1), we get

$4.51g/cm^3=\frac{6\times (47.87g/mol)}{(6.022\times 10^{23}mol^{-1}) \times V}$

$V=1.06\times 10^{-22}cm^3$

Now we have to calculate the atomic radius for Ti.

Formula used :

$V=6R^2c\sqrt{3}$

Given:

c/a ratio = 1.669 that means, c = 1.669 a

Now put (c = 1.669 a) and (a = 2R) in this formula, we get:

$V=6R^2\times (1.669a)\sqrt{3}$

$V=6R^2\times (1.669\times 2R)\sqrt{3}$

$V=(1.669)\times (12\sqrt{3})R^3$

Now put all the given values in this formula, we get:

$1.06\times 10^{-22}cm^3=(1.669)\times (12\sqrt{3})R^3$

$R=1.45\times 10^{-8}cm$

Therefore, the atomic radius for Ti is, $1.45\times 10^{-8}cm$

3 0

3 years ago