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slega [8]
3 years ago
10

Select all that apply. which of the following are density labels? kg/l g/m g/ml cm^3/g

Chemistry
2 answers:
Strike441 [17]3 years ago
4 0
I would say that only the only density label is cm^3/g
valkas [14]3 years ago
4 0

The density is defined as the ratio of mass of a substance to volume of substance

density = mass / volume

the mass of any substance is measured either in grams or Kilo grams

while volume is measured as : cm^3 or mL or L or m^3

Thus correct answer will be

kg/l

g/ml

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Please help me with my 8th grade chemistry, I'm really confused
saveliy_v [14]

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The answer would be C, in both chemical and physical change, because everything is made up of matter. So if you have a chemical or physical change, that doesent change the total mass of your product ever.

Explanation:

7 0
3 years ago
Sometimes you are given a measured amount of two or more reactants and asked how much product can be formed. Explain what steps
Westkost [7]

Explanation:

To solve this problem, follow these steps;

  • Obtain a balanced equation of the reaction and familiarize with the reactants and products.
  • Find the number of moles of the reacting species since they are the known matter in terms of quantity.
  • From the number of moles, determine the limiting reactant.
  • The limiting reactant is the one given in short supply.
  • Such reactant determines the extent of the reaction.
  • Compare the moles of this specie to that of the products using the balanced equation.
  • Obtain the mole of the desired product and find the mass or desired quantity.
  • simply work from the known specie to the unknown

learn more:

Number of moles brainly.com/question/13064292

#learnwithBrainly

4 0
4 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔
Vilka [71]

Answer : The correct option is, (C) 1.1

Solution :  Given,

Initial moles of N_2O_4 = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

The given equilibrium reaction is,

                           N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initially                      c                 0

At equilibrium   (c-c\alpha)           2c\alpha

The expression of K_c will be,

K_c=\frac{[NO_2]^2}{[N_2O_4]}

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

where,

\alpha = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}

K_c=1.066\aprrox 1.1

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0
3 years ago
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