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Inessa [10]
2 years ago
6

Explain why the production of a gas in this experiment is or is not different from the formation of water vapor in the heating c

urve experiment.
Chemistry
1 answer:
den301095 [7]2 years ago
6 0

The production of a gas is different because unlike the heating curve, there was no heat applied to this experiment, and the experiment was a chemical reaction.

<h3>What is a Heating curve?</h3>

This shows the relationship between the temperature of the heating system and the air temperature in the outer environment.

The chemical reaction emitted heat as against the other which involved heat application which is why water formation in this curve was different.

Read more about Heating curve here brainly.com/question/14094062

#SPJ1

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What does the law of conservation of mass state?
katrin [286]

Answer:

matter cannot be created nor destroyed

Explanation:

8 0
3 years ago
Read 2 more answers
Cu20(s) + C(s) - 2Cu(s) + CO(g)
Romashka-Z-Leto [24]

Answer:

That means Cu2O is limiting reagent and C is excess reagent

Explanation:

Based on the reaction, 1 mole of Cu2O reacts per mole of C. The ratio of reaction is 1:1.

To solve this question we need to convert the mass of each reactant to moles. The reactant with the lower amount of moles is limiting reactant and the excess reactant is the reactant with the higher number of moles.

<em>Moles Cu2O -Molar mass: 143.09 g/mol-</em>

114.2g Cu2O * (1mol / 143.09g) = 0.798 moles Cu2O

<em>Moles C -Molar mass: 12.01g/mol-</em>

11.1g C * (1mol / 12.01g) = 0.924 moles C

<h3>That means Cu2O is limiting reagent and C is excess reagent</h3>

<em> </em>

<em />

6 0
3 years ago
Is Radon an elementary substance or a compound?
vladimir1956 [14]

Answer:elementary substance

Explanation:

8 0
3 years ago
Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of
never [62]

Answer:

−153.1 J / (K mol)

Explanation:

Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of the species at that temperature are: Sºm (Hg,liq) = 76.02 J / (K mol) ; Sºm (Cl2,g) = 223.07 J / (K mol) ; Sºm (HgCl2,s) = 146.0 J / (K mol)

Hg(liq) + Cl2(g) → HgCl2(s)

Given that;

The standard molar entropies of the species at that temperature are:

Sºm (Hg,liq) = 76.02 J / (K mol) ;

Sºm (Cl2,g) = 223.07 J / (K mol) ;

Sºm (HgCl2,s) = 146.0 J / (K mol)

The standard molar entropies of reaction = Sºm[products] - Sºm [ reactants]

= 146.0 J / (K mol) – [76.02 J / (K mol) +223.07 J / (K mol) ]

= -153.09 J / (K mol)

= or -153.1 J / (K mol)

Hence the answer is  −153.1 J / (K mol)

5 0
3 years ago
in the following reaction 10.5 moles of chromium are added to 6.8 moles of oxygen find limiting an excess reactants and answer t
drek231 [11]

Answer:

If 3.4 moles of carbon dioxide (CO2) form at the end of the reaction, 1.7 moles of oxygen gas

(02) and 3.4 moles of carbon monoxide (CO) entered the reaction

Explanation:

The rule of three or is a way of solving problems of proportionality between three known

values and an unknown value, establishing a relationship of proportionality between all of

them. That is, what is intended with it is to find the fourth term of a proportion knowing the

other three. Remember that proportionality is a constant relationship or ratio between different

magnitudes.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so

does the other (or when one magnitude decreases, so does the other), the direct rule of three

must be applied. To solve a direct rule of three, the following formula must be followed:

ab

cx

So

C*b

T =

a

In this case, it is possible to use the reaction stoichiometry of the reaction (that is, the

relationship between the amount of reagents and products in a chemical reaction) and the rule

of three to calculate the quantities of moles requested:

.

,

If by stoichiometry 2 moles of CO2 are formed by 1 mol of O2, if 3.4 moles of CO2 are

formed, how many moles of O2 are needed?

molesofo2 = 3.42

=

moles of O2=1.7 moles

• If by stoichiometry 2 moles of CO2 are formed by 2 moles of CO, if 3.4 moles of CO2 are

formed, how many moles of CO are needed?

molesof CO =

=

3.4*2

2

moles of CO=3.4

8 0
2 years ago
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