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Arada [10]
3 years ago
14

Balance the equation and identify the type of reaction for ? P4(s) + ? Ca(s) → ? Ca3P2(s). 1. 2; 6; 2 — decomposition 2. 2; 6; 2

— displacement 3. 1; 6; 2 — displacement 4. 1; 6; 2 — synthesis 5. 2; 6; 2 — synthesis 6. 1; 6; 2 — decomposition
Chemistry
1 answer:
pishuonlain [190]3 years ago
5 0

Answer:

4. 1; 6; 2 — synthesis

Explanation:

<u>Decomposition reaction </u>is defined as the reaction in which a single large substance breaks down into two or more smaller substances.

AB\rightarrow A+B

<u>Single displacement </u>reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.

The reactivity of metal is determined by a series known as reactivity series. The metals lying above in the series are more reactive than the metals which lie below in the series.

A+BC\rightarrow AC+B

<u>Synthesis reaction</u> is defined as the reaction in which smaller substances combine in their elemental state to form a larger substance.

A+B\rightarrow AB

The unbalanced combustion reaction is shown below as:-

P_4+Ca\rightarrow Ca_3P_2

On the left hand side,  

There are 4 phosphorus atoms and 1 calcium atom

On the right hand side,  

There are 2 phosphorus atoms and 3 calcium atoms

Thus,  

Right side, Ca_3P_2 must be multiplied by 2 to balance phosphorus.

Left side, Ca is multiplied by 6 so to balance the whole reaction.

Thus, the balanced reaction is:-

P_4+6Ca\rightarrow 2Ca_3P_2

Thus, answer:- 4. 1; 6; 2 — synthesis

You might be interested in
How many atoms are in 5.00g of carbon
melisa1 [442]

Answer:

2.51 x 1023 atoms C

Explanation:

MW C = 12.011

5g x (1mol/12.011g) =0.416 mol

0.416 mol x (6.022x10^23) = 2.51 x 1023 atoms C

5 0
3 years ago
7. A gas has a volume of 300 mL at 300 mm Hg. What will its volume be if the pressure is changed to 500 mm Hg?​
USPshnik [31]

Answer:

The volume is

<h2>180 mL</h2>

Explanation:

In order to solve for the volume we use the formula for Boyle's law which is

<h3>P _{1}  V _{1} = P _{2}V _{2}</h3>

where

P1 is the initial pressure

V1 is the initial volume

P2 is the final pressure

V2 is the final volume

Since we are finding the final volume we are finding V2

Making V2 the subject we have

<h3>V _{2}  = \frac{P _{1}  V _{1}}{P _{2}  }</h3>

From the question

P1 = 300 mmHg

V1 = 300 mL

P2 = 500 mmHg

Substitute the values into the above formula and solve for the final volume obtained

That's

<h3>V _{2} =  \frac{300 \times 300}{500}  \\  =  \frac{90000}{500}  \\  =  \frac{900}{5}</h3>

We have the final answer as

<h3>180 mL</h3>

Hope this helps you

7 0
4 years ago
Consider the reaction of solid aluminum iodide and potassium metal to form solid potassium iodide and aluminum metal.The balance
ratelena [41]

Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

67.8% = 30.25 / Theoretical yield

67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

8 0
3 years ago
PI3Cl2 is a nonpolar molecule. Based on this information, determine the I−P−I bond angle, the Cl−P−Cl bond angle, and the I−P−Cl
alexdok [17]

Answer: Yes I-P-Cl = 90

Explanation:

This is because the angle formed between I-P-Cl is perpendicular hence the angle is 90°

3 0
3 years ago
An ethylene gas torch requires 300 L of gas at 0.8 atm. What will be the pressure of the gas if ethylene is supplied by a 200.0
Vlad [161]

Answer:

1.2 atm

Explanation:

Given data

  • Volume of the gas in the tank (V₁): 200.0 L
  • Pressure of ethylene gas in the tank (P₁): ?
  • Volume of the gas in the torch (V₂): 300 L
  • Pressure of the gas in the torch (P₂): 0.8 atm

If we consider ethylene gas to be an ideal gas, we can find the pressure of ethylene gas in the tank using Boyle's law.

P_1 \times V_1 = P_2 \times V_2\\P_1 = \frac{P_2 \times V_2}{V_1} = \frac{0.8atm \times 300L}{200.0L} = 1.2 atm

3 0
3 years ago
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