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2 years ago

I need help please Which option is chemical change?

Chemistry

1 answer:

MrRa [10]2 years ago

8 0

Answer:

Getting coated with tarnish

Explanation:

Silver, the original substance is combining with a new substance tarnish to make a whole new chemical.

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What does it mean to classify?

pashok25 [27]

Answer:

Explanation:

to separate objects or ideas into group based on ways they are alike

3 0

2 years ago

What would happen if the Earth was tilted more than its current tilt of 23.5°? <br><br> HELP ASAP!!

Oduvanchick [21]

Answer:

Wouldn't the Earth's atmosphere be moving too fast that it eventually breaks out?

Explanation:

Do NOT trust me.

7 0

3 years ago

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0

2 years ago

Complete the equation to show how hydrofluoric acid, HF, acts as a Brønsted-Lowry acid in water. Balanced equation:

shepuryov [24]

Answer : The balanced equation is,

$HF+H_2O\rightleftharpoons H_3O^++F^-$

Explanation :

According to the Bronsted Lowry concept, Bronsted Lowry-acid is a substance that donates one or more hydrogen ion in a reaction and Bronsted Lowry-base is a substance that accepts one or more hydrogen ion in a reaction.

The balanced chemical reaction will be,

$HF+H_2O\rightleftharpoons H_3O^++F^-$

In this reaction, hydrofluoric acid is act as a Bronsted Lowry-acid because it donate one hydrogen ion to $H_2O$ and $H_3O^+$ is act as a Bronsted Lowry-base because it accept one hydrogen ion from $HF$ .

8 0

3 years ago

I need help with this question

VARVARA [1.3K]

Answer:

2.25×10¯³ mm.

Explanation:

From the question given above, we obtained the following information:

Diameter in micrometer = 2.25 μm

Diameter in millimetre (mm) =?

Next we shall convert 2.25 μm to metre (m). This can be obtained as follow:

1 μm = 1×10¯⁶ m

Therefore,

2.25 μm = 2.25 μm / 1 μm × 1×10¯⁶ m

2.25 μm = 2.25×10¯⁶ m

Finally, we shall convert 2.25×10¯⁶ m to millimetre (mm) as follow:

1 m = 1000 mm

Therefore,

2.25×10¯⁶ m = 2.25×10¯⁶ m /1 m × 1000 mm

2.25×10¯⁶ m = 2.25×10¯³ mm

Therefore, 2.25 μm is equivalent to 2.25×10¯³ mm.

6 0

3 years ago