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olga nikolaevna [1]
3 years ago
7

If there is a bar of iron and you heat it up will it weigh the same as it did before

Chemistry
1 answer:
mario62 [17]3 years ago
8 0

No cause ur burning of particles and ions it will weigh more than an identical cold one

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Combustion vapor-air mixtures are flammable over a limited range of concentrations. The minimum volume % of vapor that gives a c
DochEvi [55]
Combustion equation of n-hexane:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

a)
Assuming we have 100 moles of air,
Oxygen = 20.9 moles
n-hexane required = 20.9/19 x 2
= 2.2 moles

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LFL n-hexane = 1.1% 

b)
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6 0
3 years ago
Read 2 more answers
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!

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3 years ago
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Which geologic time did humans exist​
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<h2><u>PLEASE MARK BRAINLIEST!</u></h2>

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