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2 years ago

Mass in grams of 6.25 mol of copper (II) nitrate?

2 answers:

8 0

Cu = 63.546
N= 14.001 g/mol
O= 15.999 g/mol * 3 = 47.997

Copper (II) Nitrate has a MW of 125.544 g/mol

6.25 x 125.544

= 784.65 <--- is your answer, if there were was a multiple choice or not :)

Margaret [11]2 years ago

3 0

Answer:

Mass of 6.25 moles of copper(II) nitrate = 1025 grams

Explanation:

Let us calculate the molar mass of Copper (II) nitrate

Molecular formula = Ca(NO₃)₂

Atomic mass of Ca = 40 g/mol

Atomic mass of N =14 g/mol

Atomic mass of O = 16 g/mol

molar mass of Ca(NO₃)₂ = 40 + (14 X2) +( 6X16) = 164 g/mol

$moles=\frac{mass}{molarmass}$

therefore

mass = moles X molar mass

mass = 6.25 X 164 = 1025 grams

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What is the molality of a solution in which 3.0 moles of NaCl is dissolved in 1.5 Kg of water?

Nesterboy [21]

Density H2O = 1g/cm³
1,5 kg H2O = 1500g = 1500cm³ (1dm³ = 1000cm³)

3moles of NaCl-----in---------1500cm³ H2O
x moles of NaCl ----in--------1000cm³ H2O
x = 2moles of NaCl

answer: 2 mol/dm³

5 0

2 years ago

Calculate the mass in grams of each of the following amounts: 1.002 mol of chromium 4.08 x 10-8 mol of neon

Pepsi [2]

Answer:

$Mass_{chromium}=52.1\ g$

$Mass_{neon}=8.23\times 10^{-7}\ g$

Explanation:

Calculation of the mass of chromium as:-

Moles = 1.002 moles

Molar mass of chromium = 51.9961 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$1.002\ mol= \frac{Mass}{51.9961\ g/mol}$

$Mass_{chromium}=1.002\times 51.9961\ g = 52.1\ g$

Calculation of the mass of neon as:-

Moles = $4.08\times 10^{-8}$ moles

Molar mass of neon = 20.1797 g/mol

Thus,

$1.002\ mol= \frac{Mass}{20.1797\ g/mol}$

$Mass_{neon}=4.08\times 10^{-8}\times 20.1797\ g = 8.23\times 10^{-7}\ g$

6 0

2 years ago

495 cm3 of oxygen gas and 877 cm3 of nitrogen gas, both at 25.0 C and 114.7 kpa, are injected into an evacuated 536 cm3 flask. F

I am Lyosha [343]

Answer:

Total pressure of the flask is 2.8999 atm.

Explanation:

Given data:

Volume of oxygen (O2) gas= 495 cm3

= 0.495 L (1 cm³ = 1 mL = 0.001 L)

Volume of nitrogen (N2) gas = 877 cm3

= 0.877 L (1 cm³ = 1 mL = 0.001 L)

volume of falsk = 536 cm3

= 0.536 L (1 cm³ = 1 mL = 0.001 L)

Temperature = 25 °C

T = (25°C + 273.15) K

= 298.15 K

Pressure = 114.7 kPa

= 114.700 Pa

Pressure (torr) = 114,700 / 101325

= 1.132 atm

Formula:

PV=nRT (ideal gas equation)

P = pressure

V = volume

R (gas constnt)= 0.0821 L.atm/K.mol

T = temperature

n = number of moles for both gases

Solution:

Firstly we will find the number of moles for oxygen and nitrogen gas.

For Oxygen:

n = PV / RT

n = 1.132 atm × 0.495 L / 0.0821 L.atm/K.mol × 298.15 K

= 0.560 / 24.47

= 0.0229 moles

For Nitrogen:

n = PV / RT

n = 1.132 atm × 0.877 / 0.0821 L.atm/K.mol × 298.15 K

n = 0.992 / 24.47

= 0.0406

Total moles = moles for oxygen gas + moles for nitrogen gas

= 0.0229 moles + 0.0406 moles

n = 0.0635 moles

Now put the values in formula

PV=nRT

P = nRT / V

P = 0.0635 × 0.0821 L.atm/K.mol × 298.15 K / 0.536 L

P = 1.554 / 0.536

P = 2.8999 atm

Total pressure in the flask is 2.8999 atm, while assuming the temperature constant.

7 0

2 years ago

Read 2 more answers

A person loses 2.70 lbs in two weeks how many grams did he lose?

Evgesh-ka [11]

He person lost 1224.7 grams.

6 0

3 years ago

Read 2 more answers

Consider the following chemical reaction: CO (g) + 2H2(g) ↔ CH3OH(g) At equilibrium in a particular experiment, the concentratio

AlexFokin [52]

Answer:

The equilibrium concentration of CH₃OH is 0.28 M

Explanation:

For the reaction: CO (g) + 2H₂(g) ↔ CH₃OH(g)

The equilibrium constant (Keq) is given for the following expresion:

Keq= $\frac{(CH3OH)}{(CO) x (H2)^{2}}$ =14.5

Where (CH3OH), (CO) and (H2) are the molar concentrations of each product or reactant.

We have:

(CH3OH)= ?

(CO)= 0.15 M

(H2)= 0.36 M

So, we only have to replace the concentrations in the equilibrium constant expression to obtain the missing concentration we need:

14.5= $\frac{(CH_{3}OH) }{(0.15 M) x (0.36 M) ^{2} }$

14.5 x (0.15 M) x $(0.36)^{2}$ = (CH₃OH)

0.2818 M = (CH₃OH)

6 0

2 years ago