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vlada-n [284]
3 years ago
14

How to find formula for ionic compounds.

Chemistry
1 answer:
Umnica [9.8K]3 years ago
5 0

Answer:

I am using an example for this in this table it works for any other compound as well.

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If a transverse wave travels 10 meters in 5 seconds what is its speed
matrenka [14]
2 meters per second. You do 10 divided by 5 to find your answer, which is 2.
5 0
4 years ago
A first-order reaction (A → B) has a half-life of 25 minutes. If the initial concentration of A is 0.900 M, what is the concentr
Oksanka [162]

Answer:

0.6749 M is the concentration of B after 50 minutes.

Explanation:

A → B

Half life of the reaction = t_{1/2}=25 minutes

Rate constant of the reaction = k

For first order reaction, half life and half life are related by:

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{25 min}=0.02772 min^{-1}

Initial concentration of A = [A]_o=0.900 M

Final concentration of A after 50 minutes = [A]=?

t = 50 minute

[A]=[A]_o\times e^{-kt}

[A]=0.900 M\times e^{-0.02772 min^{-1}\times 50 minutes}

[A] = 0.2251 M

The concentration of A after 50 minutes = 0.2251 M

The concentration of B after 50 minutes = 0.900 M - 0.2251 M = 0.6749 M

0.6749 M is the concentration of B after 50 minutes.

7 0
3 years ago
Iron III sulfide is classified as what type
Oxana [17]

Answer:

ionic compound

Explanation:

that is the answer if you meant what type of compound.

5 0
3 years ago
Read 2 more answers
Please solve quickly!!!
Maurinko [17]

Answer:

18 oxygen atoms

Explanation:

in order to from the 6 molecules carbon dioxide and 6 molecules of water you will have a total of 18 oxygen atoms

3 0
3 years ago
For the following reaction, 22.6 grams of nitrogen monoxide are allowed to react with 4.64 grams of hydrogen gas . nitrogen mono
antiseptic1488 [7]

Answer:

- 10.5 g of N₂

- Limiting reagent: NO

- 3.13 g of H₂ remains

Explanation:

First of all we state the reaction: 2NO(g) + 2H₂(g) → 2H₂O(l) + N₂(g)

We need to find out the limiting reactant and the excess reagent

Ratio in the reactants is 2:2. Let's convert the mass to moles:

22.6 g / 30 g/mol = 0.753 moles of NO

4.64 g / 2 g/mol = 2.32 moles of H₂

Certainly the limiting reagent is the NO and the excess reactant is the hydrogen:

- For 0.753 moles of NO, we need 0.753 moles of H₂ (we have 2.32 moles)

- For 2.32 moles of H₂, we need 2.32 moles of NO (and we don't have enough NO, because we only have 0.753 moles)

As the H₂ is the excess reagent, some moles still remains after the reaction is complete → 2.32 mol - 0.753 mol = 1.567 moles

We convert the moles to mass: 1.567 mol . 2g /1mol = 3.13 g of H₂ remains

As the NO is the limiting reagent, we can work with the equation:

We propose this rule of three: 2 moles of NO can produce 1 mol of N₂

Then, 0.753 moles of NO must produce (0.753 . 1) /2 = 0.376 moles of N₂

We convert the moles to mass 0.376 mol . 28 g / 1 mol = 10.5 g

3 0
4 years ago
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