All categories

3 years ago

What is the mass of 1.0 x 10-3 mole of Zn? 3.9 x 1022 0.065 0.000015

Chemistry

2 answers:

Luda [366]3 years ago

8 0

1.0 x 10^-3 x 65.38 = 0.0654 g Zn

denpristay [2]3 years ago

6 0

Answer:

0.06538 g

Explanation:

Moles of the element = Mass of the element/ Atomic mass of the element

Given,

Moles of Zn = 1.0 x 10⁻³ mol

Atomic mass of Zn = 65.38 amu (atomic mass unit)

Mass of Zn = moles of Zn x atomic mass of Zn = 1.0 x 10⁻³ x 65.38 = 65.38 x 10⁻³ g

Expressing 65.38 x 10⁻³as a decimal we have,

Mass of Zn = 0.06538 g

You might be interested in

How many grams of F2 gas are there in a 5.00-L cylinder at 4.00 × 10^3 mm Hg and 23°C?

GuDViN [60]

Answer:

41.17g

Explanation:

We are given the following parameters for Flourine gas(F2).

Volume = 5.00L

Pressure = 4.00× 10³mmHG

Temperature =23°c

The formula we would be applying is Ideal gas law

PV = nRT

Step 1

We find the number of moles of Flourine gas present.

T = 23°C

Converting to Kelvin

= °C + 273k

= 23°C + 273k

= 296k

V = Volume = 5.00L

R = 0.08206L.atm/mol.K

P = Pressure (in atm)

In the question, the pressure is given as 4.00 × 10³mmHg

Converting to atm(atmosphere)

1 mmHg = 0.00131579atm

4.00 × 10³ =

Cross Multiply

4.00 × 10³ × 0.00131579atm

= 5.263159 atm

The formula for number of moles =

n = PV/RT

n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K

n = 1.0834112811moles

Step 2

We calculate the mass of Flourine gas

The molar mass of Flourine gas =

F2 = 19 × 2

= 38 g/mol

Mass of Flourine gas = Molar mass of Flourine gas × No of moles

Mass = 38g/mol × 1.0834112811moles

41.169628682grams

Approximately = 41.17 grams.

3 0

3 years ago

What is the balanced NET ionic equation for the reaction when aqueous Cs₃PO₄ and aqueous AgNO₃ are mixed in solution to form sol

rusak2 [61]

Answer:

$PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)$

Explanation:

Hello!

In this case, since the net ionic equation of a chemical reaction shows up the ionic species that result from the simplification of the spectator ions, which are those at both reactants and products sides, we take into account that aqueous species ionize into ions whereas liquid, solid and gas species remain unionized. In such a way, for the reaction of cesium phosphate and silver nitrate we can write the complete molecular equation:

$Cs_3PO_4(aq)+3AgNO_3(aq)\rightarrow Ag_3PO_4(s)+3CsNO_3(aq)$