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Lelechka [254]
3 years ago
12

When lithium nitride, Li3N(s) , is treated with water, H2O(l) , ammonia, NH3(g) , is produced. Predict the formula of the gas pr

oduced when sodium phosphide, Na3P(s) , is treated with water. formula: Na_{3}P(aq) +H_{2}O(l) -> F_{3}(g) +NaOH(aq) Na3P(aq)+H2O(l)⟶PH3(g)+NaOH(aq) Write the balanced equation for the reaction of Na3P(s) with H2O(l) . Phases are optional. balanced equation: Na_{3}P +3 H _{2}O->PH_{3} +3NaOH
Chemistry
1 answer:
Ronch [10]3 years ago
7 0

<u>Answer:</u> The gas produced when sodium phosphide reacts with water is phosphine.

<u>Explanation:</u>

When sodium phosphide reacts with water molecule, it leads to the production of flammable, poisonous gas known as phosphine along with the production of sodium hydroxide.

The chemical reaction for the reaction of sodium phosphide with water follows the equation:

Na_3P(aq.)+3H_2O(l)\rightarrow PH_3(g)+3NaOH(aq.)

By Stoichiometry of the reaction:

1 mole of sodium phosphide reacts with 3 moles of water to produce 1 mole of phosphine gas and 3 moles of sodium hydroxide.

Hence, the gas produced when sodium phosphide reacts with water is phosphine.

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What volume of air is needed to burn completely 100cm of propane?
Evgesh-ka [11]

Answer:

See below

Explanation:

propane mole weight = 44 gm / mole

100 gm / 44 gm / mole = 2.27 moles

From the equation, 5 times as many moles of OXYGEN (O2)are required

   = 11.36 moles of oxygen

           at <u>STP</u> this is 254.55  liters of O2 (because 22.4 L = one mole) and

Using oxygen as 21 percent of air means that

      .21 x = 254.55 =       x = <u>1212.12 liters of air required </u>

5 0
2 years ago
Positive Deviation from Raoult's Law occurs when the vapour pressure of component is greater than what is expected in Raoult's L
8090 [49]

Answer:

All of the above.

Explanation:

In positive deviation from Raoult's  Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.

When a solution is non ideal then it shows positive or negative deviation.

Let two solutions A and B to form non- ideal solutions.let the vapour pressure  of component A is P_A and vapour pressure of component B is P_B.

P^0_A= Vapour pressure of component A in pure form

P^0_B= Vapour pressure of component B in pure form

x_A=Mole fraction of component A

x_B==Mole fraction of component B

The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.

P_A >P^0_A\cdot x_A,P_B>P^0_B\cdot x_B

Therefore, P_A+P_B >P^0_A\cdot x_A+P^0_B \cdot x_B

Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.

Hence, option a,b,c and d are true.

6 0
3 years ago
The pH of a solution is 8. Some hydrochloric acid is added to the solution. Suggest the pH of the solution after mixing. pH=....
Molodets [167]

Answer:

Probably around 6 because the ph of hydrochloric acid is 3

Explanation:

3 0
2 years ago
Which of the following would be found in both eukaryotic AND prokaryotic cells?
beks73 [17]
Nucleus ,endoplasmic reticulum
7 0
2 years ago
Read 2 more answers
93.0 mL of O2 gas is collected over water at 0.930 atm and 10.0C what would be the volume of this dry gas at STP
Dahasolnce [82]
The  volume  of  the  dry   gas  at     stp  is  calculated as follows

calculate  the  number  on  moles  by use  of   PV =nRT  where  n  is  the number  of  moles

n  is therefore  = Pv/RT
P  = 0.930  atm
R(gas  contant=  0.0821  L.atm/k.mol
V= 93ml  to  liters =  93/1000= 0.093L
T=  10  +  273.15 = 283.15k

n=  (0.930  x0.093)  /(0.0821  x283.15) =  3.  72  x10^-3  moles

At  STp    1  mole  =  22.4L
what about  3.72  x10^-3 moles

by  cross  multiplication

volume =  (3.72  x10^-3)mole  x  22.4L/  1  moles  = 0.083 L   or  83.3 Ml
3 0
3 years ago
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