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Elan Coil [88]
3 years ago
15

Which is more volatile benzene or toluene?

Chemistry
1 answer:
oee [108]3 years ago
4 0
Benzene is more volatile.
You might be interested in
What volume does 2.25g of nitrogen gas, N2, occupy at 273 Celsius and 1.02 atm​
kotykmax [81]
<h2><u>Answer:</u></h2>

0.126 Liters

<h2><u>Explanation:</u></h2>

V = mRT / mmP

First, convert the 2.25g of Nitrogen gas into moles. (m in the equation above)

2.25g x 1 mole / 28.0g = 0.08036 moles = m

28.0g = mm

Next, convert the 273 Celsius into Kelvin. (T in the equation above)

273 Celsius + 273.15 = 546.15K = T

R = 0.08206L*atm/mol*K

(Quick Note: The R changes depending on the Pressure Unit so do not use this number every time.)

Now, plug everything into the equation.

V = (0.08036)(0.08206)(546.15)/(28.0)(1.02)

V = 0.126 L

5 0
3 years ago
Given the equation Ca + H2O --&gt; Ca(OH)2 + H2 how many grams of calcium will react completely with 10.0 grams of water? *
Yuki888 [10]

Answer: 11.0 g of calcium will react with 10.0 grams of water.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of H_2O

\text{Number of moles}=\frac{10.0g}{18g/mol}=0.55moles

The balanced chemical equation is:

Ca+2H_2O\rightarrow Ca(OH)_2+H_2

According to stoichiometry :

2 moles of H_2O require = 1 mole of Ca

Thus 0.55 moles of H_2O require=\frac{1}{2}\times 0.55=0.275moles  of Ca  

Mass of Ca=moles\times {\text {Molar mass}}=0.275moles\times 40g/mol=11g

Thus 11.0 g of calcium will react with 10.0 grams of water.

3 0
3 years ago
What is the concentration of Htions at a pH = 11?
Lostsunrise [7]
Answer:
a) 1 x 10^-11 mol/L
b) 1 x 10^-6 mol/L
c) 1 x 10^-5 fewer H+ ions

Explanation

pH stands for Power of Hydrogen, the more acidic a substance is, the more H+ ions it has rendering the substance acidic. a pH of 1 means the concentration of H+ ions is 1 x 10^-1. A pH of 7 means the concentration of H+ ions is 1 x 10^-7 and so on.

10^-11 has 10^-5 more H+ ions than 10^-6

Hope this helps :)
5 0
3 years ago
Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el
givi [52]

Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains 6.022\times 10^{23} number of particles.

We know that:

Charge on 1 electron = 1.6\times 10^{-19}C

Charge on 1 mole of electrons = 1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C

Cu^{2+}+2e^-\rightarrow Cu

2\times 96500=193000C is passed to deposit = 1 mole of copper

63.5 g of copper is deposited by = 193000 C

14\times 1000g=14000g of copper is deposited by =\frac{193000}{63.5}\times 14000=42551181 C

To calculate the time required, we use the equation:

I=\frac{q}{t}

where,

I = current passed = 40.0 A

q = total charge = 42551181 C

t = time required = ?

Putting values in above equation, we get:

40.0=\frac{42551181 C}{t}\\\\t=1063779sec

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, 1063779s\times \frac{1hr}{3600s}=295hr

Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

3 0
3 years ago
Which subscripts would properly complete the formula unit Al_N_?
MakcuM [25]
I believe that it would be Al1N1.
4 0
3 years ago
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