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FromTheMoon [43]
3 years ago
7

A student using this technique finds that when putting a hot piece of metal into a calorimeter which contains 35.070g of water,

the water temperature increases from 22.4 c to 24.7
c. how much heat (q) was absorbed by the water (s.h of water = 4.184 j/g k)
Chemistry
1 answer:
Aleks04 [339]3 years ago
5 0
Q=cmΔt,

c=<span>4.184 j/(g*⁰C)
m=35.070 g
</span>Δt=t(final)-t(initial) = 24.7-22.4=2.3⁰C
<span>
Q=4.184</span> j/(g*⁰C)* 35.070 g * 2.3⁰C ≈ 337.49 J ≈ 338 J
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Which of the following are examples of electromagnetic waves?
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C

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Help I'm giving you a lot of points and do not put random or crazy stuff the real question
dybincka [34]

Answer:

Endothermic

Explanation:

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6 0
3 years ago
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Alchen [17]

Answer:

T₂  = 84.375 K

Explanation:

Given data:

Initial volume = 3.3 L

Initial pressure = 2000 torr

Initial temperature = 225 K

Final temperature = ?

Final volume = 2.75 L

Final pressure = 900 torr

Formula:

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₁V₁/T₁ = P₂V₂/T₂  

T₂  = P₂V₂ T₁ /P₁V₁

T₂  = 900 torr× 2.75 L× 225 K / 2000 torr×3.3 L

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7 0
3 years ago
A. Which reactant is the limiting reagent?
Tasya [4]

Answer:

a. Zinc is the limiting reactant.

b. m_{ZnBr_2}^{by\ Zn}=162.61gZnBr_2

c. m_{Br_2}^{leftover}=6.6g

Explanation:

Hello there!

a. In this case, when zinc metal reacts with bromine, the following chemical reaction takes place:

Zn+Br_2\rightarrow ZnBr_2

Thus, since zinc and bromine react in a 1:1 mole ratio, we can compute their reacting moles to identify the limiting reactant:

n_{Zn}=47.2g*\frac{1mol}{65.38g} =0.722molZn\\\\n_{Br_2}=122g*\frac{1mol}{159.8g} =0.763molBr_2

Thus, since zinc has the fewest moles we infer it is the limiting reactant.

b. Here, we compute the grams of zinc bromide via both reactants:

m_{ZnBr_2}^{by\ Zn}=0.722molZn*\frac{1molZnBr_3}{1molZn} *\frac{225.22gZnBr_2}{1molZnBr_2} =162.61gZnBr_2\\\\m_{ZnBr_2}^{by\ Br_2}=0.763molBr_2*\frac{1molZnBr_3}{1molBr_2} *\frac{225.22gZnBr_2}{1molZnBr_2} =171.95gZnBr_2

That is why zinc is the limiting reactant, as it yields the fewest moles of zinc bromide product.

c. Here, since just 0.722 mol of bromine would react, we compute the corresponding mass:

m_{Br_2}^{reacted}=0.722molBr_2*\frac{159.8gBr_2}{1molBr_2} =115.4gBr_2

Thus, the leftover of bromine is:

m_{Br_2}^{leftover}=122g-115.4g\\\\m_{Br_2}^{leftover}=6.6g

Best regards!

8 0
3 years ago
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