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VashaNatasha [74]
3 years ago
10

Can somebody plz help answer these true or false questions correctly thanks! (Only if u for sure know them) :)

Chemistry
1 answer:
asambeis [7]3 years ago
7 0

Answer:

1. False

2. True

3. False

4. True

5. True

Explanation:

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Consider a solution containing 0.100 m fluoride ions and 0.126 m hydrogen fluoride. the concentration of hydrogen fluoride after
Nataly_w [17]
Fluorine ions reacts with Hydrogen chloride to form more hydrogen fluoride.
Therefore, moles of HCl = 0.005 l × 0.01 M = 5 ×10^-5 moles
The initial moles of Hydrogen fluoride will be;
 = 0.0126 M× 0.0250 = 0.00315 Moles
Moles of hydrogen fluoride after the addition of HCl 
= 0.00315 + 5.0× 10^-5 = 0.0032 moles 
Therefore, the concentration of Hydrogen chloride 
            = 0.0032 moles/ 0.030 L 
            = 0.107 M
3 0
3 years ago
Read 2 more answers
How many calories of heat are necessary to raise the temperature of 319.5 g of water from 35.7 °C
rjkz [21]

20600Cal              

Explanation:

Given parameters:

Mass of water = 319.5g

Initial temperature = 35.7°C

Final temperature = 100°C

Unknown:

Calories needed to heat the water = ?

Solution:

The calories is the amount of heat added to the water. This can be determined using;

     H  =   m  c Ф

c  = specific heat capacity of water = 4.186J/g°C

   H is the amount of heat

    Ф is the change in temperature

    H = m c (Ф₂ - Ф₁)

    H = 319.5 x 4.186 x (100 - 35.7) = 85996.56J

Now;

     1kilocalorie = 4184J

     

85996.56J to kCal; \frac{85996.56}{4184}   = 20.6kCal  = 20600Cal

               

learn more:

Specific heat brainly.com/question/3032746

#learnwithBrainly

6 0
3 years ago
what temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution (specific heat capacity = 3
ivolga24 [154]

Answer:

1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.

Explanation:

Mass of ethylene glycol = m = 100 g

Specific heat capacity of ethylene glycol = c = 3.5 J/g°C

Change in temperature of ethylene glycol = ΔT

Heat loss by the ethylene glycol = Q = 350 J

Q=mc\Delta T

\Delta T=\frac{Q}{mc}=\frac{350 J}{100 g\times 3.5 J/g^oC}

ΔT = 1°C

1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.

3 0
3 years ago
Describe how oxidation and reduction involve electrons, change oxidation numbers, and combine in
Sholpan [36]

Answer:

Redox

Explanation:

Reduction is gain of electrons

oxidation is loss of electrons

3 0
3 years ago
Need boy who is at leaast 12
Dahasolnce [82]

Answer:

helloo!!!!!!! I am 12 :)

4 0
2 years ago
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