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Taya2010 [7]
3 years ago
10

Hello! Im in a unit exam right now and i really need help! Who ever answers this gets brainliest

Chemistry
1 answer:
worty [1.4K]3 years ago
6 0

low r value thats the answer nice to help u

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3 0
3 years ago
How many liters of ammonia (NH3), at 3.2 atm and 23C, must be used to produce of 2.65 grams of calcium hydride (CaH2). 6 Ca(s)
7nadin3 [17]

Answer:

The answer to your question is    V = 0.32 L

Explanation:

Data

Volume of NH₃ = ?

P = 3.2 atm

T = 23°C

mass of CaH₂ = 2.65 g

Balanced chemical reaction

               6Ca  +  2NH₃   ⇒   3CaH₂  +  Ca₃N₂

Process

1.- Convert the mass of CaH₂ to moles

-Calculate the molar mass of CaH₂

 CaH₂ = 40 + 2 = 42 g

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

                              2.65 g --------------  x

                              x = (2.65 x 1)/42

                              x = 0.063 moles

2.- Calculate the moles of NH₃

                     2 moles of NH₃ --------------- 3 moles of CaH₂

                      x                        --------------- 0.063 moles

                                x = (0.063 x 2) / 3

                                x = 0.042 moles of NH₃

3.- Convert the °C to °K

Temperature = 23°C + 273

                      = 296°K

4.- Calculate the volume of NH₃

-Use the ideal gas law

              PV = nRT

-Solve for V

                V = nRT / P

-Substitution

                V = (0.042)(0.082)(296) / 3.2

-Simplification

               V = 1.019 / 3.2

-Result

               V = 0.32 L

7 0
3 years ago
0.415 g of an unknown triprotic acid are used to make a 100.00 mL solution. Then 25.00 mL of this solution is transferred to an
Arturiano [62]

Answer:

Explanation:

Initial burette reading = 1.81 mL

final burette reading = 39.7 mL

volume of NaOH used = 39.7 - 1.81 = 37.89 mL .  

37.89 mL of .1029 M NaOH is used to neutralise triprotic acid

No of moles contained by 37.89 mL of .1029 M NaOH

= .03789 x .1029 moles

= 3.89 x 10⁻³ moles

Since acid is triprotic ,  its equivalent weight = molecular weight / 3

No of moles of triprotic acid = 3.89 x 10⁻³ / 3

= 1.30   x 10⁻³ moles .

8 0
3 years ago
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