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Paladinen [302]
2 years ago
6

At STP conditions, 0.25 mole of CO 2 (g), H 2 (g), NH 3 (g) will:

Chemistry
1 answer:
bija089 [108]2 years ago
3 0
 <span>1 mole of any substance contains 6.02 x 10^23 particles. 

Hence 0.25 mole of any has will contain 0.25 x 6.02 x 10^23 molecules of that gas

They cannot contain the same no. of atoms as 1 molecule of CO2 contains 3 atoms, 1 molecule H2 contains 2 molecules, and 1 molecule of NH3 contain 4 atoms

1 molecule of any gas will occupy 22.4 L at STP

hence 0.25 mol of all of these gases accupy the same volume

all of these molecules have different molar masses. thus their g.f.w cannot be same

Your answer is that they will occupy the same volume, and the same number of molecules</span>
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What is solubility curves​
love history [14]

Answer:

Explanation:

A solubility curve is a graph of solubility, measured in g/100 g water, against temperature in °C. Solubility curves for more than one substance are often drawn on the same graph, allowing comparisons between substances

8 0
2 years ago
Give the structure of the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane.
Fudgin [204]

the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane is 2-bromo-2,4,4-trimethylpentane.

what is free radical halogenation?

A substitution reaction in which a hydrogen atom is replaced with a halogen atom, via a free radical reaction mechanism. when this reaction is carrid out by bromine radical, it is called free radicle bromination. When bromine (Br{2}) treated with light (hν) it comes to homolytic cleavage of the Br-Br bond and give rise to bromine radicles.

free-radical bromination of 2,2,4−trimethylpentane

Bromination of an alkane includes the substitution of a bromine atom for a hydrogen atom. The following stages will be taken by 2,2,4-trimethylpentane during this reaction:

Initiation reaction:  The production of a bromine free radical requires the initiation of heat or light.

Br - Br ⇒ 2Br·

Propagation: This reaction relies heavily on hydrogen. This reaction is impossible if hydrogen is not present. Because tertiary free radicals are more stable than secondary and primary free radicals, they are favoured in this reaction.

Termination: The remaining free radical of bromide reacts with the tertiary free radical of 2,2,4-trimethylpentane to form 2-bromo-2,4,4-trimethylpentane.

the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane is 2-bromo-2,4,4-trimethylpentane.

To know more about free radical halogenation, check out:

brainly.com/question/13046867

#SPJ4

7 0
1 year ago
TRUE - FALSE
inessss [21]

Answer:

True

Explanation:

4 0
2 years ago
A syringe contains 0.65 moles of he gas that occupy 900.0 ml. what volume (in l) of gas will the syringe hold if 0.35 moles of n
Grace [21]
<span>PV=nRT= a universal constant For any condition P1V1/n1T1=R and P2V2/n2T2=R i.e P1V1/n1T1=P2V2/n2T2 Becomes V1/n1=V2/n2 rearranging and solving V2=V1X(n2/n1)= 750 mLx((0.65+0.35)/(0.65))=1200ml=1.2L...2 sig figs</span>
5 0
3 years ago
Read 2 more answers
BRAINLIESTTT ASAP! PLEASE HELP ME :)
Iteru [2.4K]

Answer:

The mass of the products left in the test tube will be less than that of the original reactants.

Explanation

The equation for the reaction is

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

 1.0          3.0              3.9             0.1

Assume you started with 1.0 g of Mg.

It will react with 3.0 g of HCl to form 3.9 g of MgCl2 and 0.1 g of H2

.

Mass of reactants = mass of products

        1.0 g + 3.0 g = 3.9 g + 0.1 g

                    4.0 g = 4.0 g

The Law of Conservation of Mass is obeyed.

However, your test tube and its contents will weigh 0.1 g less than it did before the reaction.

Does that contradict the Law of Conservation of Mass? It does not.

One of the products was the gas, hydrogen, and it escaped from the test tube. You weren't measuring all the products, so test tube and its contents weighed less than before.

7 0
2 years ago
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