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Monica [59]
3 years ago
8

Which of the following describes a limiting reactant? Question 10 options: A) The reactant that runs out last and limits how muc

h product can be made B) The reactant that is present at the smallest mass C) The reactant that runs out first and limits how much product can be made D) An inert element
Chemistry
1 answer:
miskamm [114]3 years ago
7 0

Answer:

The correct choice is option C) The reactant that runs out first and limits how much product can be made

Explanation:

The limiting reactant is a reactant in a chemical reaction that gets consumed first and thus limits how much product can be formed.

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The equation below shows the products formed when a solution of silver nitrate (AgNO3) reacts with a solution of sodium chloride
d1i1m1o1n [39]
I would say that the answer has to be C
Since there is no change in mols on both sides of the equation the mass is constant
3 0
4 years ago
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How much heat will be given off when 1500 g of water cools down by 20 C? please help and explain:)))
docker41 [41]

Answer:

126000J

Explanation:

From the question given, we obtained the following information:

M = 1500g

C = 4.2J/g°C

ΔT = 20°C

Q =?

Q= MCΔT

Q = 1500 x 4.2 x 20 =

Q = 126000J

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4 years ago
Weight and mass are essentially the same measurements.<br> True or false?
Serhud [2]

The answer to this question is False

7 0
3 years ago
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Choose all the answers that apply.
max2010maxim [7]

Answer: has properties similar to other elements in group 18, does not react readily with other elements, is part of the noble gas group

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6 0
3 years ago
Please show some work For the reaction: NO(g) + 1/2 O2(g) → NO2(g) ΔH°rxn is -114.14 kJ/mol. Calculate ΔH°f of gaseous nitrogen
uranmaximum [27]

Answer:

148.04 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

NO(g) + 1/2 O₂(g) → NO₂(g)      ΔH°rxn = -114.14 kJ/mol

We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.

ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))

ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol

ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol

ΔH°f(NO(g)) = 148.04 kJ/mol

8 0
3 years ago
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