The balanced equation for the reaction between Na and H₂SO₄ is as follows;
2Na + H₂SO₄ ---> Na₂SO₄ + H₂
Stoichiometry of Na to H₂SO₄ is 2:1
the molar concentration of H₂SO₄ is 6.0 M
This means that in 1 L of H₂SO₄ - there are 6.0 mol of H₂SO₄
Therefore in 750 mL -
![\frac{6.0 mol* 750 mL}{1000mL }](https://tex.z-dn.net/?f=%20%5Cfrac%7B6.0%20mol%2A%20750%20mL%7D%7B1000mL%20%7D%20)
Number of H₂SO₄ moles reacted - 4.5 mol
According to stoichiometry,
1 mol of H₂SO₄ reacts with 2 mol of Na
Therefore 4.5 mol reacts with 2x4.5= 9 mol of Na
Mass of Na reacted - 9 mol x 23 g/mol = 207 g of Na is required
The answer is x = 2.6 x 10^24. Hope this helps. :)
I do believe it is C. wave length <span />
Answer:
They also showed the effects of pressure on volume if temperature stayed the same
Explanation:
They also showed the effects of pressure on volume if temperature stayed the same is the experiment that will provide an evidence for Boyle's law.
Boyle's law states that "the volume of a fixed mass of a gas varies inversely as the pressure changes, if the temperature is constant".
- The law is an affirmation of what happens when there is a dynamics between pressure and volume if temperature is made constant.
- So the experiment designed to investigate this proves and shows Boyle's law.
Answer:
1.209g of MgO participates
Explanation:
In this problem, we have 0.030 moles of MgO that participates in a particular reaction.
And we are asked to solve for the mass of MgO that participates, that means, we need to convert moles to grams.
To convert moles to grams we need to use molar mass of the compound:
<em>1 atom of Mg has a molar mass of 24.3g/mol</em>
<em>1 atom of O has a molar mass of 16g/mol</em>
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That means molar mass of MgO is 24.3g/mol + 16g/mol = 40.3g/mol
And mass of 0.030 moles of MgO is:
0.030 moles MgO * (40.3g/mol) =
<h3>1.209g of MgO participates</h3>