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SOVA2 [1]
3 years ago
13

In a titration 5.0 mL of a 2.0 M NaOH aq solution exactly neutralizes 10.0 of an HCL aq solution what is the concentration of th

e HCL(aq)solution
Chemistry
1 answer:
KonstantinChe [14]3 years ago
8 0
The  concentration    of  HCl solution  is  calculated  as  follows

write  the  equation   for  reaction

NaOH  + HCl = NaCl  +  H2O

find  the  moles   NaOH  =  molarity  x volume /1000

=  5  x2/1000  =  0.01  moles

by  use  of mole  ratio  between  NaOh  to HCl  which  is 1 :1  the  the  moles of  Hcl   is  also =  0.01 moles

concentration is therefore =   moles/volume    x1000

=  0.01/10  x1000 = 1M

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OLga [1]

Answer:

Gravity is the answer.

5 0
2 years ago
Determine the percentage of carbon and hydrogen in ethane C2H6 if the molecular weight is 30.
Naddika [18.5K]

Answer:

Percentage of carbon:

{ \tt{ =  \frac{24}{30}  \times 100\%}} \\  = 80\%

Percentage of hydrogen:

{ \tt{ =  \frac{6}{30}  \times 100\%} } \\  = 20\%

8 0
3 years ago
"What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0"
Artyom0805 [142]

Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)

Answer:

-3670.33 J/K

Explanation:

Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.

Mathematically,  change of Entropy can be expressed as,

ΔS = ΔH/T ....................................... Equation 1

Where ΔS = Change of entropy, ΔH = heat change, T = temperature.

ΔH = -(Lf×m).................................... Equation 2

Note: ΔH is negative because heat is lost.

Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg

Substitute into equation

ΔH = -(3.34×10⁵×3.0)

ΔH = - 1002000 J.

But T = 0 °C = (0+273) K = 273 K.

Substitute into equation 1

ΔS = -1002000/273

ΔS = -3670.33 J/K

Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C

4 0
2 years ago
What is the lower concentration limit (vol%) at which a mixture of ethanol in air can explode?
liubo4ka [24]

Answer:

Lower explosive limit (LEL) of ethanol = 3.3%

Explanation:

In the case of alcohol, ethanol presents certain fire hazards. Its momentary flash point is 55ºF (12.9ºC), while the momentary flash point of gasoline is -45ºF (-42.8ºC), and the E85 mixture ranges between -20ºF and -4ºF (between -28 , 9ºC and -20ºC), and has a wider range of flammability limits than gasoline. For emergency response teams, this implies that during a release of the typical ethanol / gasoline mixture, the fuel can be expected to behave like gasoline: It is heavier than air - as we mentioned earlier - and can produce vapors and form flammable mixtures in the air, under most environmental conditions.

General properties and comparison with other inflambles products:

Flash point momentary Gasoline = -45 ° F

<u>Ethanol</u> = 55 ° F

E 85 = between -20º and -4º F

<u>Flammability limits </u>

Lower explosive limit (LEL) of ethanol = 3.3%

Upper Explosive Limit (UEL) = 19%

Lower explosive limit (LEL) of the mixture E 85 = 1.4%

Upper Explosive Limit (UEL) 85 = 19%

Lower explosive limit (LEL) of gasoline = 1.4%

Upper Explosive Limit (UEL) = 7.6%

They have a wider range than gasoline

4 0
2 years ago
ASAP: The main difference between the gravitational force and electrical force is that
Dahasolnce [82]

Answer:

Electrical force can pull and push

Explanation:

5 0
3 years ago
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