All categories

3 years ago

HELP ME PLEASEEE ILL MARk u the brainliest

Chemistry

2 answers:

Novay_Z [31]3 years ago

8 0

Answer:

3rd or 4th answer is correct

belka [17]3 years ago

5 0

OPTION B is the correct answer

You might be interested in

Razak uses 70 W fan and a 40 w lamp for eight hours a day, Calculate the amount of energy he used in a month. NORS Analysing

Anastasy [175]

Answer:

70 W and 40 w

there are 28 days in a month

Add those 2 together and get 110

then mutlpy by 28

3,080 amount of energy

6 0

3 years ago

Which of the following is a polar substance? O Water O Grease O Gasoline O Oil

JulijaS [17]

Water will be your answer

5 0

3 years ago

Which statement correctly describes potassium iodide, KI? There is a one-to-one ratio of potassium ions to iodide ions. Potassiu

elena-14-01-66 [18.8K]

The answer is there is a one-to-one ratio of potassium ions to iodide ions.

Explanation :

- (K) belongs to Alkali metals in group (1A) that contains (1) electron in the outermost energy level, whereas, (I) is from halogens in group (7A) that contains (7) electron in the outermost energy level.

- To achieve stability, both atoms tend to reach the nearest noble state (outermost level occupies 8 electrons). Therefore, (K) loses its outer electron and gives it to (I) which now has a completely filled outer level and an ionic bond is formed between the two.

- The valency (number of electrons lost, gained or shared) of both atoms is equal ”monovalent” which means one-to-one ratio..

5 0

3 years ago

Read 2 more answers

A chemical reaction that absorbs energy in the form of heat is considered _____.

Zinaida [17]

Endothermic reaction is the answer.

5 0

3 years ago

PLEASE HELP ME ASAP! CHEMISTRY TUTOR SEE ATTACHED

Masteriza [31]

Answer:

$\large \boxed{\text{-827.4 kJ}}$

Explanation:

We have three equations:

1. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic) + O₂(g) ⟶ SO₂(g); ∆H = -296.8 kJ

3. PbO(s) + H₂S(g) ⟶ PbS(s) + SO₂(g); ∆H = -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s) + 3O₂(g) ⟶2PbO(s) + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.

7. 2S(s, rhombic) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

6. 2H₂S(g) + O₂(g) ⟶ 2S(s) + 2H₂O(g) ; ∆H = -442.4 kJ

7. 2S(s) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

4 . 2PbS(s) + 3O₂(g) ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

$\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}$

8 0

3 years ago