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igomit [66]
2 years ago
15

The molecular geometry of the PF3 molecule is __________, and this molecule is ________.

Chemistry
1 answer:
MArishka [77]2 years ago
5 0

Answer:

C.) trigonal pyramidal, polar

Explanation:

Around the phosphorus central atom, there are 3 fluorine atoms and 1 lone pair. This makes the shape tetrahedral, but gives it a trigonal pyramidal geometry. The lone pair on the phosphorus disrupts the balance of the molecule and pulls the electrons towards it. This makes the molecule polar.

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Which example best demonstrates the relationship between temperature and pressure? bubbles forming in dough when it is heated bu
Margaret [11]
I think the correct answer from the choices listed above is the first option. The best example that demonstrates the relationship between temperature and pressure would be that <span>bubbles forming in dough when it is heated. Hope this answers the question.</span>
8 0
3 years ago
How many grams of o2(g) are needed to completely burn 70.2 g of C3H8(g)
Lubov Fominskaja [6]
First you must write a balanced chemical equation.

C3H8 + 5O2 --> 3CO2 + 4H2O

From there, we can set up the stoichiometry equation to solve.

g O2= 70.2 g C3H8 X (1 mol C3H8/44.0962g C3H8) X (5 mol O2/1 mol C3H8) X (31.998g O2/1 mol O2)

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6 0
3 years ago
How are crystals made? ​
Step2247 [10]

Explanation:

Crystals often form in nature when liquids cool and start to harden. Certain molecules in the liquid gather together as they attempt to become stable. They do this in a uniform and repeating pattern that forms the crystal. In nature, crystals can form when liquid rock, called magma, cools.

4 0
3 years ago
Read 2 more answers
While heating two different samples of water at sea level, one boils at 102°c and one boils at 99.2°c. calculate the percent err
Evgesh-ka [11]
<span>To find: Sample error in percent
 Solution:
 Formula:
((Experimental value-theoretical value)/theoretical value)*100
 where, theoretical value = 100°c
 and, experimental value = 102°c (sample 1) 99.2°c (sample 2)
 Sample error (in percentage) when boiling level was 102°c = ((102°c-100°c)/100°c)*100 = 2%
 Sample error (in percentage) when boiling level was 99.2°c = ((99.2°c-100°c)/100°c)*100 = -0.8%</span>
3 0
4 years ago
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Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces
Alchen [17]

The question is incomplete, here is the complete question:

Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces only nitrogen and water vapor, and in the liquid form it is easily transported. An industrial chemist studying this reaction fills a 5.0 L flask with 2.2 atm of ammonia gas and 2.4 atm of oxygen gas at 44.0°C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of nitrogen gas to be 0.99 atm.

Calculate the pressure equilibrium constant for the combustion of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.

<u>Answer:</u> The pressure equilibrium constant for the reaction is 32908.46

<u>Explanation:</u>

We are given

Initial partial pressure of ammonia = 2.2 atm

Initial partial pressure of oxygen gas = 2.4 atm

Equilibrium partial pressure of nitrogen gas = 0.99 atm

The chemical equation for the reaction of ammonia and oxygen gas follows:

                    4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)

<u>Initial:</u>               2.2          2.4

<u>At eqllm:</u>        2.2-4x      2.4-3x         2x        6x

Evaluating the value of 'x':  

\Rightarrow 2x=0.99\\\\x=0.495

So, equilibrium partial pressure of ammonia = (2.2 - 4x) = [2.2 - 4(0.495)] = 0.22 atm

Equilibrium partial pressure of oxygen gas = (2.4 - 3x) = [2.4 - 3(0.495)] = 0.915 atm

Equilibrium partial pressure of water vapor = 6x = (6 × 0.495) = 1.98 atm

The expression of K_p for above equation follows:

K_p=\frac{(p_{N_2})^2\times (p_{H_2O})^6}{(p_{NH_3})^4\times (p_{O_2})^3}  

Putting values in above equation, we get:

K_p=\frac{(0.99)^2\times (1.98)^6}{(0.22)^4\times (0.915)^3}\\\\K_p=32908.46

Hence, the pressure equilibrium constant for the reaction is 32908.46

5 0
3 years ago
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