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IRISSAK [1]
3 years ago
11

Which substance would require the most energy to heat 10.00 g of it by 10.0°C?

Chemistry
1 answer:
Leviafan [203]3 years ago
8 0

Answer:

a

Explanation:

a

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bezimeni [28]

Explanation:

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Describe the three main sub?atomic particles. Include mass (with units), charge and location for each.
Afina-wow [57]
Electron - negligible mass, negative charge, orbits the nucleus
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5 0
3 years ago
Consider the first-order reaction shown here. the yellow spheres in the pictures to the right represent the reactant,
Luda [366]
The rate constant of the reaction K we can get it from this formula:

K=㏑2/ t1/2 and when we have this given (missing in question):
that we have one jar is labeled t = 0 S and has 16 yellow spheres inside and the jar beside it labeled t= 10 and has 8 yellow spheres and 8 blue spheres and the yellow spheres represent the reactants A and the blue represent the products B

So when after 10 s  and we were having 16 yellow spheres as reactants and becomes 8 yellow and 8 blue spheres as products so it decays to the half amount so we can consider T1/2 = 10 s
a) by substitution in K formula:
∴ K = ㏑2 / 10 = 0.069 
The amount of A (the reactants) after N half lives = Ao / 2^n
b) so no.of yellow spheres after 20 s (2 half-lives) =  16/2^2 = 4
and the blue spheres = Ao - no.of yellow spheres left = 16 - 4 = 12
c) The no.of yellow spheres after 30 s (3 half-lives) = 16/2^3 = 2
and the blue spheres = 16 - 2 = 14

8 0
3 years ago
NEED HELP ASAP NOT DIFFICULT
Burka [1]
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.

Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
5 0
2 years ago
Read 2 more answers
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