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2 years ago

14

help me balance & find how many grams of water would be produced if 20.0 liters of O2 gas were burned at a temperature of -1

0.O*c & a pressure of 1.3 atm

1 answer:

vova2212 [387]2 years ago

6 0

I have finished the balanced equation part. Give me a little more time for part B.

Edit: Found a way so here it is as well.

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Balance this equation C2H4 + O2 + CO2 + H2O

LenKa [72]

Answer: C₂H₄+3 O₂= 2 CO₂+ 2 H₂O

Explanation:

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2 years ago

at what temperature will you find more potassium nitrate(kno3) in 100 g of H2O . In other words at which temperature will you ha

Oksana_A [137]

The answer to this question would be: <span>85c
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3 years ago

What do you look at to tell the reactivity of an atom?

tia_tia [17]

Answer:

The number of electrons in the outermost shell of an atom determines its reactivity. Noble gases have low reactivity because they have full electron shells. Halogens are highly reactive because they readily gain an electron to fill their outermost shell.

Explanation:

I hope this helped!

8 0

2 years ago

All of the following institutions awards certificates except: a.) Community college b.) liberal arts college c. Junior college d

Serjik [45]

Except B, Liberal Arts College

3 0

3 years ago

Read 2 more answers

When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II)

lesya692 [45]

Answer: The mass of lead deposited on the cathode of the battery is 1.523 g.

Explanation:

Given: Current = 62.0 A

Time = 23.0 sec

Formula used to calculate charge is as follows.

$Q = I \times t$

where,

Q = charge

I = current

t = time

Substitute the values into above formula as follows.

$Q = I \times t\\= 62.0 A \times 23.0 sec\\= 1426 C$

It is known that 1 mole of a substance tends to deposit a charge of 96500 C. Therefore, number of moles obtained by 1426 C of charge is as follows.

$Moles = \frac{1426 C}{96500 C/mol}\\= 0.0147 mol$

The oxidation state of Pb in $PbSO_{4}$ is 2. So, moles deposited by Pb is as follows.

$Moles of Pb = \frac{0.0147}{2}\\= 0.00735 mol$

It is known that molar mass of lead (Pb) is 207.2 g/mol. Now, mass of lead is calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\ 0.00735 = \frac{mass}{207.2 g/mol}\\mass = 1.523 g$

Thus, we can conclude that the mass of lead deposited on the cathode of the battery is 1.523 g.

3 0

2 years ago