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Anuta_ua [19.1K]
3 years ago
12

Iron (II) chloride Covalent or ionic

Chemistry
1 answer:
ExtremeBDS [4]3 years ago
3 0

Answer:

ionic

Explanation:

Iron(II) chloride, also known as ferrous chloride, is the chemical compound of formula FeCl2. It is a paramagnetic solid with a high melting point. The compound is white, but typical samples are often off-white

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Question 2
SVETLANKA909090 [29]

D- Physical

Explanation:

A physical property is anything that has characteristics associated with a change in it's chemical composition

3 0
3 years ago
CHEMISTRY HELP<br> Once the following equation is balanced, what is the correct coefficient for Z₂?
yawa3891 [41]

Answer:

The coefficient of Z₂ is 1.

Explanation:

From the question given above:

X + ZY —> XY + Z₂

Next, we shall balance the equation to obtain the coefficient of Z₂. This can be obtained as follow:

X + ZY —> XY + Z₂

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

X + 2ZY —> XY + Z₂

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

X + 2ZY —> 2XY + Z₂

Now, we have 1 atom of X on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of X as shown below:

2X + 2ZY —> 2XY + Z₂

Now the equation is balanced.

Thus, the coefficient of Z₂ is 1.

8 0
3 years ago
At 700 K, the reaction2SO2(g) + O2(g) 2SO3(g)has the equilibrium constant Kc = 4.3 x 106, and the following concentrations are p
dedylja [7]

Explanation:

The given reaction is as follows.

       2SO_{2} + O_{2}(g) \rightarrow 2SO_{3}(g)

Value of equilibrium constant is given as K_{c} = 4.3 \times 10^{6}[/tex].

Concentration of given species is [SO_2] = 0.010 M; [SO_3] = 10.M; [O_2] = 0.010 M.

Formula for experimental value of equilibrium constant (Q) is as follows.

             Q = \frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}

Putting the given concentration as follows.

              Q = \frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}

             Q = \frac{(10)^{2}}{(0.010)^{2}(0.010)}

              Q = 10^{8}

It is known that when Q > K_{eq}, then reaction moves in the backward direction.

When Q < K_{eq}, then reaction moves in the forward direction.

When Q = K_{eq}, then reaction is at equilibrium.

As, for the given reaction Q > K_{eq} then it means reaction moves in the backward direction.

Thus, we can conclude that the reaction is moving in the backward direction, that is, right to left to reach the equilibrium.

5 0
3 years ago
Addition of 50. J to a 10.0-g sample of a metal will cause the temperature of a metal to rise from 25ºC to 35ºC. The specific he
Snowcat [4.5K]

Answer:

b) C = 0.50 J/(g°C)

Explanation:

  • Q = mCΔT

∴ Q = 50 J

∴ m = 10.0 g

∴ ΔT = 35 - 25 = 10 °C

specific heat (C) :

⇒ C = Q / mΔT

⇒ C = 50 J / (10.0 g)(10 °C)

⇒ C = 0.50 J/(g°C)

7 0
3 years ago
Draw all four products obtained when 2-ethyl-3-methyl-1,3-cyclohexadiene is treated with HBr at room temperature and show the me
LenKa [72]

Answer:

See explanation below

Explanation:

In this case we have reaction of addition. In this case a diene reacting with an acid as HBr. This reaction is known as Hydrohalogenation, and, as we have a diene, this kind of reaction can be done as 1,4 addition. Which means that the reaction will be undergoing with an adition in the carbon 1, and carbon 4.

At room temperature we can expect that this reaction can be done in thermodynamic conditions, Now, as the problem states that is forming 4 products, we can expect products of a 1,2 addition too. This product can be formed if the reaction is taking place in the most stable carbocation, and then, by resonance, we can expect the 1,4 product too.

Now, the HBr can be attacked by the double bond of the first position, giving two possible products or by the double bond of the third position giving the other two products. These products are all possible, obviously the most stable will be the major of all of them, but the other three are perfectly possible. One product is formed without doing much, and the other by resonance. Same happens with the other double bond.

In the picture below, you have the mechanism for all the 4 products.

Hope this helps

5 0
3 years ago
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