Answer:
The answer is 2.20 M
Explanation:
This is because ammonia has a pH of 11.8 and if you take 14-11.8 it equals 2.2 so the answer is 2.20 M
1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.
How to calculate number of atoms?
The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.
However, the number of moles of oxygen in glycine can be calculated using the following expression:
Molar mass of C₂H5O2N = 75.07g/mol
Mass of oxygen in glycine = 32g/mol
Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine
Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles
Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms
Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.
Learn more about number of atoms at: brainly.com/question/8834373
#SPJ1
Hello.
If you wanted to harvest cranberries you would need to use flotation.
Have a nice day.
0.370 mol metal oxide = 55.45 g
<span>1 mol = 55.45/0.370 = 149.86 g </span>
<span>in 1 mol there are 3 mol O = 16 * 3 = 48 g of O </span>
<span>there is 48/149.86 * 100% O in the sample </span>
<span>the sample has 48/149.86 * 0.370 = 0.119 g O</span>
Answer is: <span>- delta G.
</span>The change in Gibbs free energy (ΔG), at constant temperature and pressure, is: <span>ΔG=ΔH−TΔS.
</span>ΔH<span> is the change in enthalpy.
</span>ΔS is change in entropy.
T is temperature of the system.
When ΔG is negative, a reaction (<span>occurs without the addition of external energy)</span><span> will be spontaneous (</span>exergonic).
