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Rom4ik [11]
2 years ago
7

The volume of a sample of oxygen is 300.0 mL, when the pressure is 1.00 atm and the temperature is 300K. At what pressure will t

he volume be 1000.0 mL and a temperature of 500K?
Chemistry
1 answer:
Brut [27]2 years ago
4 0

Answer:

The new pressure is 0.5 atm

Explanation:

Step 1: Data given

Volume of oxygen = 300 mL = 0.300 L

Pressure = 1.00 atm

Temperature = 300 K

The volume increases to 1000mL = 1.00 L

The temperature increases to 500 K

Step 2: Calculate the new pressure

(P1*V1)/T1 = (P2*V2)/T2

⇒with P1 = the initial pressure = 1.00 atm

⇒with V1 = the initial volume = 0.300 L

⇒with T1 = the initial temperature = 300 K

⇒with P2 = the new pressure = TO BE DETERMINED

⇒with V2 = the increased volume = 1.00 L

⇒with T2 = the increased temperature = 500 K

(1.00 atm* 0.300 L)/300 K = (P2 * 1.00L) / 500 K

P2 = (1.00 *0.300 * 500) / (300 *1.00)

P2 = 0.5 atm

The new pressure is 0.5 atm

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A) What is the maximum number of grams of nickel bromide that can be produced from the reaction of 67.8 g of nickel with 37.3 g
svetoff [14.1K]

Answer:

The answer to your question is a) 51.07 g of NiBr₂   b) Nickel, 54 g

Explanation:

Data

mass of NiBr₂ = ?

mass if Ni = 67.8 g

mass of Br = 37.3 g

Balanced chemical reaction

                Ni  +  Br₂   ⇒   NiBr₂

Process

1.- Find the atomic mass of the reactants and the molar mass of the product

Ni = 59 g

Br = 79.9 x 2 = 159.8 g

NiBr₂ = 59 + 159.8 = 218.8 g

2.- Find the limiting reactant

theoretical yield  Ni/Br₂ = 59/159.8 = 0.369

experimental yield Ni/Br₂ = 67.8/37.3 = 1.81

The limiting reactant is Bromine because the experimental yield was lower than the theoretical yield.

3.- Calculate the mass of NiBr₂

                    159.8 g of Br₂ --------------- 218.8 g of NiBr₂

                      37.3 g of Br₂ --------------  x

                          x = (37.3 x 218.8) / 159.8

                          x = 8161.24/159.8

                          x = 51.07 g of NiBr₂

4.- Find the excess reactant

The excess reactant is Nickel

                59 g of Ni ---------------- 159.8 g of Br₂

                  x               ----------------  37.3 g of Br₂

                            x = (37.3 x 59)/159.8

                            x = 2200.7/159.8

                            x = 13.77 g of Ni

Excess Ni = 67.8 - 13.77

                 = 54 g

5 0
3 years ago
NO LINKS RIGHT ANSWERS ONLY!!!!
Mumz [18]
I think the answer is D
6 0
3 years ago
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7.Calculate the volume of nitrogen that reacts with 12dm3 of hydrogen with the volume of both gases measured at rtp:
marta [7]

Answer:

4.03dm³

Explanation:

The reaction expression is given as:

       3H₂   +   N₂   →    2NH₃  

  Volume of hydrogen  = 12dm³  

AT rtp:

             1 mole of gas occupies volume of 22.4dm³  

             x mole of hydrogen will occupy a volume of 12dm³

     Number of moles of hydrogen  = \frac{12}{22.4}   = 0.54mole

From the balanced reaction equation:

            3 mole of hydrogen gas combines with 1 mole of Nitrogen gas

         0.54 mole of hydrogen as will therefore combine with \frac{0.54}{3}   = 0.18moles of nitrogen gas

Since ;

                     1 mole of gas occupies a volume of 22.4dm³

               0.18moles of Nitrogen gas will occupy 0.18 x 22.4  = 4.03dm³

6 0
3 years ago
What would cause the pressure of a gas to increase?​
IrinaK [193]

Answer: The pressure of a gas will increase when there is a decrease in the volume of the gas.

Explanation: according to Boyle's law, the volume of a gas will decrease when the pressure is increased at constant temperature and vice versa.

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Bromine has the following electron configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5. categorize the electrons in each. Answer for video: The video player is loaded.

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