Nickel has a face-centered cubic unit cell with an edge length of 352.4 pm. Calculate the density of nickel

How many grams of chlorine gas are present in a 150. liter cylinder of chlorine held at a pressure of 1.00 atm and 0. °C? Group

OlgaM077 [116]

Answer:

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

P= 1.00 atm
V= 150 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 0 C= 273 K

Replacing:

1.00 atm* 150 L= n*0.08206 $\frac{atm*L}{mol*K}$ *273 K

Solving:

$n=\frac{1.00 atm* 150 L}{0.08206 \frac{atm*L}{mol*K}*273 K}$

n= 6.69 moles

Being Cl= 35.45 g/mole, the molar mass of chlorine gas is:

Cl₂=2*35.45 g/mole= 70.9 g/mole

So if 1 mole has 70.9 grams, 6.69 moles of the gas, how much mass does it have?

$mass=\frac{6.69 moles*70.9 grams}{1 mole}$

mass= 474.321 grams ≅ 474 grams

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

4 0

3 years ago

Convert 1.05 x 1026 molecules Ni to number of moles.

vodka [1.7K]

Answer:

1077.3 molecular in 1.05*1026 molecules

4 0

3 years ago

Where was copper discovered? What country? pLeAse halp yall

Andrews [41]

Copper was discovered in North Africa.

3 0

3 years ago

Determination of Vitamin C Concentration by Redox Titration Pre-laboratory Questions 1. A solution is prepared by dissolving 2.0

Ierofanga [76]

Answer:

M = 0.011

Half reactions for reduction and oxidation:

oxidation:

HC12H6Cl2O2N + NH+ +2e- = HC12H8Cl2O2N

reduction:

C6H8O6 = C6H6O6 + 2H+ + 2e-

Explanation:

Molarity of the ascorbic acid solution:

Ascorbic acid molar mass = 173.12 g/mol

Then you have to find the mols:

mol = 2g/176,12 =0.011 mol Ascorbic acid.

Use the molarity equation:

Molarity (M) = 0.011 mol / 1 L = 0.011 M

Reactions for the reduction of DCPIP by ascorbic acid:

These reactions have to be balanced by the oxide reduction method.

Because it involves the transfer of electrons from ascorbic acid to the DCPIP.

Remember: when is happens a reduction, also happens oxidation.

To see the reactions watch the images attached.

The first one has the calculus for the molarity of the ascorbic acid solution.

The second one has the reactions for the reduction and oxidation of the compounds.

3 0

4 years ago

12. How many moles of atoms are present in 154 g of Na2O?

tia_tia [17]

Answer:

$\boxed {\boxed {\sf A. \ 2.48 \ mol \ Na_2O}}$

Explanation:

We are asked to convert an amount in grams to moles. To do this, we use the molar mass. This is the number of grams in one mole of a substance. It is the same value numerically as the atomic mass on the Periodic Table, however the units are grams per mole, not atomic mass units.

Look up the molar masses for the individual elements.

Sodium (Na): 22.9897693 g/mol
Oxygen (O): 15.999 g/mol

Look back at the formula: Na₂O. Notice there is a subscript of 2 after sodium. This means there are 2 atoms of sodium in every molecule, so we have to multiply sodium's molar mass by 2 before adding oxygen's.

Na₂O: 2(22.9897693 g/mol)+ 15.999 g/mol = 61.9785386 g/mol

Set up a ratio using the molar mass.

$\frac {61.9785386 \ g \ Na_2O}{1 \ mol \ Na_2O}$

Multiply by the given number of grams.

$154 \ g \ Na_2O*\frac {61.9785386 \ g \ Na_2O}{1 \ mol \ Na_2O}$

Flip the ratio so the grams of sodium oxide can cancel each other out.

$154 \ g \ Na_2O*\frac {1 \ mol \ Na_2O}{61.9785386 \ g \ Na_2O}$

$154 *\frac {1 \ mol \ Na_2O}{61.9785386 }$

$\frac {154}{61.9785386 } \ mol \ Na_2O$

$2.48473106141 \ mol \ Na_2O$

The original measurement of grams given has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place.

2.48473106141

The 4 in the thousandth place tells us to leave the 8.

$2.48 \ mol \ Na_2O$

There are 2.48 moles of sodium oxide in 154 grams, so choice A is correct.

6 0

3 years ago