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2 years ago

Is this any better??

Chemistry

1 answer:

Anna11 [10]2 years ago

3 0

I believe there is an error or incorrect word for the first question, but all you have to do to solve for these questions is to make use of the formula

D = m/v, solve for the correct variable by doing the opposite operation.

Divide - multiply

Vice versa.

Then solve for the asked variable.

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A clear, pure liquid sample is brought into the lab and exposed to an electrical current. Different gases are produced on each o

aleksklad [387]

It would be a compound.

4 0

3 years ago

What's the definition of ions

Masja [62]

An atom or molecule with a net electric charge due to the loss or gain of one or more electrons. :)

3 0

3 years ago

Read 2 more answers

What is the concentration of hydronium ion in a 0.121 M HCl solution?

kogti [31]

Answer:

C) 0.121 M

Explanation:

HCl + H₂O = H₃O⁺ + OH⁻

.121M .121M

HCl is a strong acid . It will dissociate almost 100 % so the concentration of acid and hydronium ion formed will be equal . It is to be noted that hydronium ion is formed due to association of H⁺ and H₂O . H⁺ is formed due to ionisation of HCl .

So concentrtion of hydronium ion ( H₃O⁺ ) will be .121 M.

7 0

3 years ago

Draw the product formed when the compound shown below undergoes a reaction with hcl in ch2cl2.

Taya2010 [7]

The product formed when HCl and CH2Cl2 react is CH4
H
H-C- H methane structure
H

HCl react with CH2Cl2 to form methane (CH4) and chlorine gas(Cl2)
that is,

2HCl(g) + CH2Cl2(l) = CH4 (g) +2Cl2 (g)

6 0

3 years ago

What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su

anastassius [24]

Answer:

$V=27992L=28.00m^3$

Explanation:

Hello,

In this case, the combustion of methane is shown below:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

$n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4$

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

$PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3$

Best regards.

8 0

3 years ago