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2 years ago

What of the following could be classified as matter

Chemistry

1 answer:

Sholpan [36]2 years ago

7 0

Answer:

Desk

Water

Cloud

Helium

Explanation:

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Which product(s) would be obtained by the dehydration of 2-heptanol? of 2-methyl-l-cyclohexanol?

Komok [63]

Hello)
1)CH3-CH(OH)-СН2-СН2-СН2-СН2-СН3---(H2SO4)--›CH3-CH=CH-CH2-CH2-CH2-CH3+H2O
2)2-methyl-l-cyclohexanol---(h2so4)--›CH2=C(CH3)-CH2-CH2-CH2-CH2-CH3+H2O

5 0

3 years ago

The molar solubility of zns is 1.6 Ã 10-12 m in pure water. Calculate the ksp for zns.

Oksi-84 [34.3K]

Answer:

The solubility product of zinc sulfide is $2.56\times 10^{-24}$ .

Explanation:

Solubility of the zinc sulfide = $S=1.6\times 10^{-12}$

$ZnS\rightleftharpoons Zn^{2+}+S^{2-}$

S S

The expression of solubility product is given by :

$K_{sp}=[Zn^{2+}]\times [S^{2-}]$

$K_{sp}=S\times S=S^2$

$=1.6\times 10^{-12}\times 1.6\times 10^{-12}=2.56\times 10^{-24}$

The solubility product of zinc sulfide is $2.56\times 10^{-24}$ .

5 0

3 years ago

A sample of water (88 grams) had a temperature change of 6.0 ℃. What was the heat change of this sample? (q = m c ΔT, c = 4.18 J

jeka94

Answer:

$Q=2209J$

Explanation:

Hello!

In this case, since the heat involved during a heating process is computed in terms of mass, specific heat and temperature change as shown below:

$Q=mC\Delta T$

Thus, since the heated mass of water was 88 g, the specific heat of water is 4.184 J/g°C and the temperature change is 6.0 °C, we can compute the heat as shown below:

$Q=88g*4.184\frac{J}{g\°C}*6.0\°C \\\\Q=2209J$

Best regards!

7 0

2 years ago

The equilibrium constant for the reaction of fluorine gas with bromine gas at 300 K is 54.7 and the reaction is: Br2(g) + F2(g)

alisha [4.7K]

Answer:

The correct answer is 0.024 M

Explanation:

First we use an ICE table:

Br₂(g) + F₂(g) ⇔ 2 BrF(g)

I 0.111 M 0.111 M 0

C -x -x 2 x

E 0.111 -x 0.111-x 2x

Then, we replace the concentrations of reactants and products in the Kc expression as follows:

Kc= $\frac{[BrF ]^{2} }{[ F_{2} ][Br_{2} ]}$

Kc= $\frac{(2x)^{2} }{(0.111-x)(0.111-x)}$

54.7= $\frac{4x^{2} }{(0.111-x)^{2} }$

We can take the square root of each side of the equation and we obtain:

7.395= $\frac{2x}{(0.111-x)}$

0.111(7.395) - 7.395x= 2x

0.82 - 7.395x= 2x

0.82= 2x + 7.395x

⇒ x= 0.087

From the x value we can obtain the concentrations in the equilibrium:

[F₂]= [Br₂]= 0.111 -x= 0.111 - 0.087= 0.024 M

[BrF]= 2x= 2 x (0.087)= 0.174 M

So, the concentration of fluorine (F₂) at equilibrium is 0.024 M.

8 0

3 years ago

The solution starts with increased

marta [7]

Answer:

Increasing the Molarity by Volume. Determine the number of moles of solute in a given solution by dividing the number of grams of solute by its molecular mass.

Explanation:

8 0

3 years ago