All categories

2 years ago

Write the chemical formula for iridium(III) nitride?

Chemistry

1 answer:

lubasha [3.4K]2 years ago

3 0

Answer:

Ir(NO2)3

Explanation:

it's Molar Mass is 330.2335 if you need that too

You might be interested in

How many mg does an 830 kg sample contain?

Aleksandr-060686 [28]

Answer:

A sample of 830 kg would contain 830000000 mg

Explanation:

3 0

2 years ago

Which would be a more practical source of electricity for a car: a wet cell or a dry cell? Why?

Finger [1]

Answer:

wet

Explanation:

4 0

2 years ago

Three mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,

8_murik_8 [283]

Answer:

(a). 46,666.7 g/mol; 78,571.4 g/mol

(b). 86950g/mol; 46,666.7 g/mol.

(c). 86950g/mol; 43,333.33 g/mol

Explanation:

So, we are given the molar masses for the three samples as: 10,000, 30,000 and 100,000 g mol−1.

Thus, the equal number of molecule in each sample = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

The average molar mass = [ ( 10,000)^2 + (30,000)^2 + 100,000)^2] ÷ 10,000 + 30,000 + 100,000 = 78,571. 4 g/mol.

(b). The equal masses of each sample = 3/[ ( 1/ 10,000) + (1/30,000 ) + (1/100,000) ] = 20930.23 g/mol.

Average molar mass = ( 10,000 + 30,000 + 100,000 ) / 3 = 46,666.7 g/mol.

(c). Equal masses of the two samples = (0.145 × 10,000) + (0.855 × 100,000)/ 0.145 + 0.855 = 86950g/mol.

The weight average molar mass = 1.7 + 10,000 + 100,000/ 1.7 + 1 = 43,333.33 g/mol.

6 0

2 years ago

When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capac

STatiana [176]

Answer:

$\Delta _{comb}H=-2,093\frac{kJ}{mol}$

Explanation:

Hello!

In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

$Q_{rxn}+Q_{cal}=0$

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

$Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ$

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

$n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\ \Delta _{comb}H=-2,093\frac{kJ}{mol}$

Best regards!

4 0

1 year ago

When 100 ml of 1.0 M Na3PO4 is mixed with 100 ml of 1.0 M AgNO3,a yellow precipitate forms and Ag+ becomes negligibly small. Whi

jok3333 [9.3K]

Answer:

Option A

Explanation:

Number of millimoles of Na3PO4 = 1 × 100 = 100

Number of millimoles of AgNO3 = 1 × 100 = 100

When 1 mole of Na3PO4 is dissociated we get 3 moles of sodium ions and 1 mole of phosphate ion

When 1 mole of AgNO3 is dissociated, we get 1 mole of Ag+ and 1 mole of NO3-

As Ag+ concentration is negligible, the dissociated Ag+ ion must have form the precipitate with phosphate ion and as number of moles of Ag+ and phosphate ion are same, therefore the concentration of phosphate ion must be negligible

Here as 100 millimoles of Na3PO4 is there, we get 300 millimoles of Na+ and 100 millimoles of PO43-

And as 100 millimoles of AgNO3 is there, we get 100 millimoles of Ag+ and 100 millimoles of NO3-

∴ Increasing order of concentration will be PO43- < NO3- < Na+

4 0

2 years ago

Read 2 more answers