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3 years ago

Increase the mass of the red ball. What happens to the force on the blue ball? What happens to the force on

Chemistry

1 answer:

gregori [183]3 years ago

5 0

Answer:

i really dont know because i dont have anything to look at to know that answer.

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You are planting flowers in large pots. You plan to plant flowers in 10 pots. 15 cups of potting soil will fill one pot. A bag o

dimaraw [331]

Answer:

I will need six (6) bags of potting soil

Explanation:

Since you plan on planting in 10 pots and need 15 cups of potting soil per pot, the total amount of potting soil you need (in cups) is 10 X 15 = 150 cups of potting soil.

We have that a bag contains 25 sups. to get the number of bags needed, we have to divide 150 by 25. This will give us 150 / 25 = 6 bags.

Therefore, I will need six (6) bags of potting soil

3 0

3 years ago

What is the change in atomic number when an atom emits a beta particle?

salantis [7]

When an atom emits a beta particle from the nucleus, the nucleus only has one more proton and one less neutron and this will make the atomic mass number remains unchanged while the atomic number increases by 1.

I hope this answer helps you!

4 0

3 years ago

Read 2 more answers

The colour imparted to a flame by calcium ion?

emmainna [20.7K]

The flame goes an Orange-Red colour.

6 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

Gaseous hydrogen iodide is placed in a closed container at 425∘C, where it partially decomposes to hydrogen and iodine: 2HI(g)⇌H

murzikaleks [220]

Answer:

0.0184

Explanation:

Let's consider the following reaction at equilibrium.

2 HI(g) ⇌ H₂(g) + I₂(g)

The concentration equilibrium constant (Kc) is equal to the product of the concentration of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients.

Kc = [H₂] × [I₂] / [HI]²

Kc = (4.78 × 10⁻⁴) × (4.78 × 10⁻⁴) / (3.52 × 10⁻³)²

Kc = 0.0184

5 0

3 years ago