Answer:
0.79 g
Explanation:
Let's introduce a strategy needed to solve any similar problem like this:
- Apply the mass conservation law (assuming that this reaction goes 100 % to completion): the total mass of the reactants should be equal to the total mass of the products.
Based on the mass conservation law, we need to identify the reactants first. Our only reactant is sodium bicarbonate, so the total mass of the reactants is:
We have two products formed, sodium carbonate and carbonic acid. This implies that the total mass of the products is:
Apply the law of mass conservation:
Substitute the given variables:
Rearrange for the mass of carbonic acid:
The elements in each group have the same number of electrons in the outer orbital. Or also called valence electrons. Khan academy has a great video online explaining why this happens. (It only happens for main group elements). Here is a link (sorry you can’t click it in Brainly) https://www.khanacademy.org/science/chemistry/periodic-table/copy-of-periodic-table-of-elements/v/periodic-table-valence-electrons. Feel free to message me for a better explanation, I would explain now but I’m not sure how much you know about this. If you know how to write an electron configuration you can see how all the electron configurations for the same group (not the transitional metals only the main groups) have the same number of valence electrons. I hope that helped, sorry I was vague about the explanation :)
<span>1. the pressure of a gas over a solvent is increase
</span> Gas solubility decreases"<span>
2. the partial pressure of an anesthetic gas is increased
</span> " gas solubility does not change"<span>
3. air in blood a diver descends 10 M and pressure increases by 1 atm
</span> Gas solubility decreases"<span>
4. the temp is increase
</span>"gas solubility increases"<span>
5. O2 the temp of a body of water rises.
</span>"gas solubility increases"
Answer:
They will have the same ear shape since their proteins connect in the same ways to determine ear shape.
Explanation:
I majored in Chemistry
