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o-na [289]
3 years ago
8

What is the volume of 0.250 m hydrochloric acid required to react completely with 20.0 ml of 0.250 m ca(oh)2?

Chemistry
1 answer:
Soloha48 [4]3 years ago
5 0
The balanced equation for the above neutralisation reaction is as follows;
Ca(OH)₂ + 2HCl ----> CaCl₂ + 2H₂O
Stoichiometry of Ca(OH)₂ to HCl is 1:2
number of Ca(OH)₂ moles reacted - 0.250 mol/L x 20.0 x 10⁻³ L = 5.00 x 10⁻³ mol
according to molar ratio of 1:2
number of HCl moles required = 2 x number of Ca(OH)₂ moles reacted
number of HCl moles = 5.00 x 10⁻³ x 2 = 10.0 x 10⁻³ mol
molarity of HCl solution - 0.250 M
there are 0.250 mol  in volume of 1 L
therefore 10.0 x 10⁻³ mol in - 10.0 x 10⁻³ mol  / 0.250 mol/L = 40.0 mL 
40.0 mL of 0.250 M HCl is required
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Which property of a substance is not altered by a physical change?
antiseptic1488 [7]
C. Composition

is the answer
3 0
3 years ago
Certain compound contains 7.3% carbon, 4.5% hydrogen, 36.4% oxygen, and 31.8% nitrogen. It’s reality molecular mass is 176.0. Fi
erastovalidia [21]

Answer:

The answer to your question is:

Explanation:

Data

carbon        7.3%          =     7.3g

hydrogen    4.5%         =      4.5g

oxygen       36.4%         =     36.4 g

nitrogen     31.8%         =     31.8 g

Now

For carbon

                    12 g --------------------1 mol

                    7.3 g     -------------     x

                       x = 7.3/12 = 0.608 mol

For hydrogen

                 1 g   --------------------  1 mol

                 4.5 g  ------------------    x

                   x = 4.5 mol

For oxygen

             16 g ------------------- 1 mol

             36.4 g ----------------    x

             x = 2.28 mol

For nitrogen

              14 g   ----------------   1 mol

              31.8 g ---------------    x

             x = 2.27 mol

Now divide by the lowest result, the is 0.608 from carbon

carbon              0.608/0.608 = 1

hydrogen           4.5/ 0.608 = 7.4

oxygen              2.28/0.608 = 3.75

nitrogen             2.27/0.608 = 3.73

Empirical formula = CH₇O₄N₄

     

6 0
3 years ago
A gas at 1.5 atm had pits ressure decreased to 0.50 atm producing a new volume of 750 ml what was its original volume
Vlada [557]

Answer:

The original volume of the gas is 0.001 mL

Explanation:

This easy excersise can be solved by the law for gases, about pressure and volume; the volume of a gas is inversely proportional to the pressure it exerts.

We can propose the rule by this formula:

P₁  / V₁ = P₂ / V₂

We replace data given: 1.50 atm / V₁ = 0.50 atm / 750 mL

As the rule says, that volume is inversely proportional, and the pressure was decreased, volume must be lower than 750 mL.

1.5atm / (0.5 atm / 750mL) = V₁

V₁ = 0.001 mL

6 0
3 years ago
If the value for ΔS is postive, and the value for ΔH is negative, thr reaction will be
dmitriy555 [2]

It follows that the reaction is spontaneous at high temperatures Option A.

<h3>What is ΔS ?</h3>

The term ΔS is referred to as the change in the entropy of the system. Now recall that entropy is defined as the degree of disorderliness in a system. If a system is highly disorderly then it means that it has a high entropy. Also, ΔH has to do with the heat change that accompanies a reaction.

We know that both the entropy and the heat change can both either be positive or negative. Now we know that the equation ΔG = ΔH - TΔS can be used to ascertain whether or not a reaction will be spontaneous. If the result is negative, then the reaction will be spontaneous.

As such, when then it follows that the reaction is spontaneous at high temperatures Option A.

Learn more about spontaneous reaction:brainly.com/question/13790391

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6 0
1 year ago
A person tries to heat up her bath water by adding 5.0 l of water at 80c to 60 l of water at 30c. what is the final temperature
Sophie [7]
We can calculate the final temperature from this formula :

when Tf = (V1* T1) +(V2* T2) / (V1+ V2)

when V1 is the first volume of water = 5 L 

and V2 is the second  volume of water = 60 L

and T1 is the first temperature of water in Kelvin = 80 °C +273 = 353 K

and T2 is the second temperature of water in Kelvin =  30°C + 273= 303 K

and Tf is the final temperature of water in Kelvin 

so, by substitution:

Tf = (5 L * 353 K ) + ( 60 L * 303 K) / ( 5 L + 60 L)

     = 1765 + 18180 / 65 L

     = 306 K
     = 306 -273 = 33° C

8 0
3 years ago
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