Answer:
Mass of the salt: 105.6g of KCl.
Mass water: 958.9g of water.
Molality: 1.478m.
Explanation:
<em>Mass of the salt:</em>
In 1L, there are 1.417 moles. In grams:
1.417 moles KCl * (74.54g / mol) = 105.6g of KCl
<em>Mass of the water:</em>
We can determine the mass of solution (Mass of water + mass KCl) by multiplication of the voluome (1L and density 1064.5g/L), thus:
1L * (1064.5g / L) = 1064.5g - Mass solution.
Mass water = 1064.5g - 105.6g = 958.9g of water
<em>Molality:</em>
Moles KCl = 1.417 moles KCl.
kg Water = 958.9g = 0.9589kg.
Molality = 1.417mol / 0.9589kg = 1.478m
Not only for plants, but also for every organism and every life form nitrogen is important because it poses as catalyst that supports different is chemical reactions of growth. N is filled with chlorophyll that actually gives you the answer on the question as chlorophyll allow plant to perform photosynthesis by absorbing energy from light that leads to the growth. So, it is important because it provides <span>photosynthesis.</span>
pH decreases as the hydrogen ion concentration increases.
<u>Explanation:</u>
When there is a decrease in pH, that is pH decreases from 6 to 3 then the acidity increases.
That is the pH is between 1 to 7 then it is acidic
When the pH is 7 then it is neutral
When the pH is between 7 to 14 then it is basic
As the H⁺ ion concentration increases, then the pH value decreases, here pH decreases from 6 to 3.
So the concentration of Hydrogen ion increases, pH decreases.
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
(1) Ocean to Continent
(2)Continent to Continent
(3)Ocean to Ocean
are the three sub types of convergent plate boundaries.