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2 years ago

What is sodium bicarbonate an example of? a buffer an acid a liquid a base

Chemistry

1 answer:

ludmilkaskok [199]2 years ago

3 0

Answer:

base

Explanation:

carbonic acid - H2CO3 - is a weak acid. Therefore, HCO3 itself is its conjugate base. The Na(sodium) ion is neutral.

This means that NaHCO3 is a base. (a weak one)

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What is the weight in grams of 3.36 × 10Æ³ molecules of copper sulfate (CuSOÁ)?

spin [16.1K]

Molar mass ( CuSO₄) = 159.609 g/mol

159.609 g ----------------- 6.02 x 10²³ molecules
? g ------------------ 3.36 x 10²³ molecules

mass = ( 3.36 x10²³) x 159.609 / 6.02 x 10²³

mass = 5.36 x 10²⁴ / 6.02 x 10²³

mass = 8.90 g

hope this helps!

6 0

3 years ago

List the names of the elements that make up the molecule?

garik1379 [7]

Answer:i nitrogen, oxygen, hydrogen

Explanation: " Molecules of most elements are made up of only one of atom of that element" well if you search it up

6 0

3 years ago

Read 2 more answers

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago

No caso da pessoa ser um atleta profissional,essa demanda de energia pode subir até 5400Kcal. Transforme essa medida de energia

ddd [48]

Answer:

If you speak Any English i Think I will be able to help you Los Amigo

Explanation:

6 0

2 years ago

Once again if anyone could help, I would really appreciate your help! Thank you ❤️

Nikolay [14]

Answer:

Explanation:

I explained how to do it on your other problem so look there :)

6 0

3 years ago