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3 years ago

What is the correct measure of density

Chemistry

2 answers:

Darya [45]3 years ago

4 0

Dividing the mass by the volume.

kifflom [539]3 years ago

4 0

Answer: A: Mass/Volume will be your answer.

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What is the volume of a 5M solution with that has 16.5 moles?

kenny6666 [7]

Molarity can be used to calculate the volume of solvent or the amount of solute. The relationship between two solutions with the same amount of moles of solute can be represented by the formula c1V1 = c2V2, where c is concentration and V is volume.

7 0

3 years ago

A sample of pure NO2 is heated to 335 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+

vodka [1.7K]

The equilibrium constant for the reaction is 0.00662

Explanation:

The balanced chemical equation is :

2NO2(g)⇌2NO(g)+O2(g

At t=t 1-2x ⇔ 2x + x moles

The ideal gas law equation will be used here

PV=nRT

here n= $\frac{w}{W}$ = $\frac{w}{V}$ = density

P = $\frac{density RT}{M}$ density is 0.525g/L, temperature= 608.15 K, P = 0.750 atm

putting the values in reaction

0.75 = $\frac{0.525 x 0.0821 x 608.15 }{M}$

M = 34.61

to calculate the Kc

Kc= $\frac{ [NO] [O2]}{NO2}$

$\frac{1-2x}{1+x}$ x M NO2 + $\frac{2x}{1+x}$ M NO+ $\frac{x}{1+x}$ M O2

Putting the values as molecular weight of NO2, NO,O2

$\frac{46(1-2x) +30(2x)+32x}{1+x}$

34.61= $\frac{46}{1+x}$

x= 0.33

Kc= $\frac{4x^2)x}{1-2x^2}$

putting the values in the above equation

Kc = 0.00662

5 0

2 years ago

Consider a mystery compound having the formula MxTy. If the compound is not an acid, if it contains only two elements, and if M

valkas [14]

Answer:

d.) It is a binary molecular compound.

Explanation:

The compound in question has a formula $M{x}T_{y}$ . The compound is not acidic in nature and the element 'M' is not a metal. This shows that the compound does not contain any metal. Based on the definition of a binary molecular compound as a compound comprising elements that are not metals. Therefore, the compound is obviously a binary molecular compound.