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Lady bird [3.3K]
3 years ago
7

When dissolved in water, which compound is generally considered to be an arrhenius acid? when dissolved in water, which compound

is generally considered to be an arrhenius acid? ch3ch2oh na2co3 naoh ch3co2h?
Chemistry
1 answer:
Contact [7]3 years ago
6 0
Answer is: <span>ch3co2h; acetic acid.
</span><span>An Arrhenius acid is a substance that dissociates in water to form hydrogen ions or protons. 
</span>Acetic acid dissociate in aqueous solution to form hydrogen ions (H⁺) and acetic anion (CH₃COO⁻).
<span>Balanced chemical reaction (dissociation):
CH</span>₃COOH(aq) ⇄ CH₃COO⁻(aq) + H⁺(aq).

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Name an ionic compound or a covalent compound, explain why it is ionic or covalent
stellarik [79]

Answer:

IOnic compound is NAcL cuz in the periodic table eft are metal and right are non metal makes it ionic (different)

Explanation:

7 0
3 years ago
All molecules are compounds<br> True or false
Harrizon [31]

False (All compounds are molecules, but not all molecules are compounds. That is because a molecule can be made up of two atoms of the same kind, as when two oxygen atoms bind together to make an oxygen molecule. However, all compounds are made up of two or more different types of atoms.)

4 0
3 years ago
What is the wavelength of spectral line resulted from the electron transition from n=3 to n=2 in a hydrogen atom.
MakcuM [25]

Answer:

\lambda=6.56x10^{-7}m

Explanation:

Hello there!

In this case, it possible to use the Rydberg equation in order to calculate the wavelength for this transition from n=3 to n=2 as shown below:

\frac{1}{\lambda} =R_H(\frac{1}{n_f^2}-\frac{1}{n_i^2} )

Thus, we plug in the corresponding energy levels and the Rydberg constant to obtain:

\frac{1}{\lambda} =10973731.6m^{-1}(\frac{1}{2^2}-\frac{1}{3^2} )\\\\\frac{1}{\lambda} =1524129.4m^{-1}\\\\\lambda=\frac{1}{1524129.4m^{-1}} \\\\\lambda=6.56x10^{-7}m

Best regards!

5 0
4 years ago
What are the answer??
MrMuchimi

Answer:?

Explanation:None

4 0
3 years ago
Read the given equation. 2Na + 2H2O → 2NaOH + H2
Crank

Answer:

B) 12.9 grams.

Explanation:

How many moles of molecules in that 6.30 L of H₂?

The volume of one mole of an ideal gas at STP (0 °C, 1 atm) is 22.4 liters.

(The volume of that one mole of gas at STP will be 22.7 liters if STP is defined as 0 °C and 10⁵ Pa).

\displaystyle n(\text{H}_2) = \frac{6.30\;\text{L}}{22.4\;\text{L}\cdot\text{mol}^{-1}} = 0.28125\;\text{mol}.

How many moles of Na will be needed?

The coefficient in front of Na in the equation is twice the coefficient in front of H₂. It takes two moles of Na to produce one mole of H₂.

n(\text{Na}) = 2\;n(\text{H}_2) = 2\times 0.28125= 0.5625\;\text{mol}.

What's the mass of that many Na atoms?

Refer to a modern periodic table. The molar mass of ₁₁Na is 22.990. The mass of one mole of Na atoms is 22.990 gram. The mass of 0.5625 moles of Na atoms will be

m = n\cdot M = 0.5625 \times 22.990 = 12.9\;\text{g}.

(2 sig. fig. as in the volume of the H₂ gas.)

6 0
3 years ago
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