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crimeas [40]
3 years ago
14

Question 1 of 6

Chemistry
1 answer:
Juli2301 [7.4K]3 years ago
8 0

Answer:

Option A. 2, 3, 2

Explanation:

We'll begin by balancing the equation. This can be achieved by doing the following:

Fe + Cl2 —> FeCl3

There are 2 atoms of Cl on the left side and 3 atoms on the right side. It can be balance by putting 3 in front of Cl2 and 2 in front of FeCl3 as shown below:

Fe + 3Cl2 —> 2FeCl3

There are 2 atoms of Fe on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Fe as shown below:

2Fe + 3Cl2 —> 2FeCl3

Now the equation is balanced.

The coefficients are : 2, 3, 2

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HELP!! An unknown solution has a pH reading of 8.4. Which statement is true about this solution?
yarga [219]

Answer:

D- a weak base

Explanation:

The ph scale goes from 1-10 a solution over 7ph is classified as basic. A solution thats 8.4 is only 1.4 over 7pH, making it a weak basic solution. An example of a strong base would be a solution with a pH of 9.2 (for example).

8 0
3 years ago
What is the mass, in grams of a pure iron cube that has a volume of 4.20cm^3
Dmitriy789 [7]
d_{Fe}=7874\frac{kg}{m^{3}}=7,874\frac{g}{cm^{3}}\\
V=4,2cm^{3}\\\\
m=dV=7,784\frac{g}{cm^{3}}*4,2cm^{3}\approx 32,7g
6 0
3 years ago
The decomposition of ammonia is: 2 NH3(g) ⇌ N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammoni
ValentinkaMS [17]

Answer:

"6.7\times 10^{-4} \ atm" is the right answer.

Explanation:

Given:

Partial pressure of N_2,

= 0.20 atm

Partial pressure of H_2,

= 0.15 atm

K_p = 1.5\times 10^3 at 400^{\circ} C

As we know,

⇒ K_p = \frac{pN_2\times pH_2^3}{pNH_3^2}

By putting the values, we get

    1.5\times 10^3=\frac{0.20\times (0.15)^3}{pNH_3^2}

        pNH_3^2 = \frac{0.000675}{1.5\times 10^3}

                    =6.7\times 10^{-4} \ atm

                   

3 0
2 years ago
The half-life of cesium-137 is 30 years. Suppose we have a 200-mg sample. (a) Find the mass that remains after t years.
scZoUnD [109]

Given:

Half life(t^ 1/2) :30 years

A0( initial mass of the substance): 200 mg.

Now we know that

A= A0/ [2 ^ (t/√t)]

Where A is the mass that remains after t years.

A0 is the initial mass

t is the time

t^1/2 is the half life

Substituting the given values in the above equation we get

A= [200/ 2^(t/30) ] mg


Thus the mass remaining after t years is [200/ 2^(t/30) ] mg

5 0
3 years ago
This is for a study guide, I can't figure it out!
miskamm [114]
<span>The answer is "D" where the number of collisions per unit area is reduced by one-half. Drawing back on the piston means the volume is increased. The pressure is reduced. There are fewer collisions when the pressure is reduced.</span>
3 0
3 years ago
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