All categories

2 years ago

PLZ I NEEEEEEEDDDDDD HELP

Chemistry

2 answers:

Mazyrski [523]2 years ago

4 0

Answer:

Explanation:

1. __1___ Fe + __1___ S → __1___ FeS. 2. __1___ H2 + __1___ Cl2 → __2___ HCl. 3. __2___ Mg + __1___ O2 → __2___ MgO. 4. __1___ O2 + __2___ H2

MrRa [10]2 years ago

3 0

Answer:

2h+02=h20

Explanation:

2 in front of h on left side

2 in front of h on right side

You might be interested in

Ms. Sharkey wanted to see which tennis shoes made her run the fastest. She bought

Svetach [21]

Answer:

to compare the data she has got?

7 0

3 years ago

ICl has a higher boiling point than Br2. What is the best explanation for this?

Flura [38]

<span>The best answer is B. ICl experiences induced dipole-induced dipole interactions. Both iodine and chlorine belongs to the same group of the periodic table. Electronegativity decreases as you go down a group therefore Cl will have a greater attraction with the bond it forms with another atom. Dipole-dipole interactions form between I and Cl. For the Br2 molecule, no dipole occurs because they are two identical atoms. Therefore we will be expecting ICl will have a higher boiling point due to higher binding energy it forms.</span>

7 0

3 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

3 years ago

What is the transmittance of 35.3%

andrey2020 [161]

This might be wrong,but im pretty sure its 0.10 m

8 0

3 years ago

The temperature in a town started out at 55 degrees. By the end of the day the temperature dropped to -6 degrees.

garik1379 [7]

Answer:

The choice of the answer is fourth option that is -61 degrees.

Therefore the temperature drop is -61°Centigrade.

Explanation:

Given:

The temperature in a town started out at 55 degrees

Start temperature = 55°Centigrade. (Initial temperature)

End of the Day = -6°Centigrade. (Final temperature)

To Find:

How far did the temperature drop?

Solution:

We will have,

$\textrm{Temperature drop}=\textrm{Final temperature}-\textrm{Initial temperature}$

Substituting the above values in it we get

$\textrm{Temperature drop}=-6-55\\\\\therefore \textrm{Temperature drop}=-61\° centigrade$

Therefore the temperature drop is -61°Centigrade.

4 0

3 years ago