We are given that the concentration of NaOH is 0.0003 M and are asked to calculate the pH
We know that NaOH dissociates by the following reaction:
NaOH → Na⁺ + OH⁻
Which means that one mole of NaOH produces one mole of OH⁻ ion, which is what we care about since the pH is affected only by the concentration of H⁺ and OH⁻ ions
Now that we know that one mole of NaOH produces one mole of OH⁻, 0.0003M NaOH will produce 0.0003M OH⁻
Concentration of OH⁻ (also written as [OH⁻]) = 3 * 10⁻⁴
<u>pOH of the solution:</u>
pOH = -log[OH⁻] = -log(3 * 10⁻⁴)
pOH = -0.477 + 4
pOH = 3.523
<u>pH of the solution:</u>
We know that the sum of pH and pOH of a solution is 14
pH + pOH = 14
pH + 3.523 = 14 [subtracting 3.523 from both sides]
pH = 10.477
Given data Atomic mass of Ra= 226g/mol
no. of moles =1.0/226g/mol =0.04424moles
no. of atoms in 0.044moles
no. of atoms =no. of moles x avogadro's number
= 0.044x 6.022 x10^23 = 0.264968 x 10^22
If 10^15 atoms of Ra produce 1,373*10^4 atoms of<u> Rn per second</u> then 2,66 *10^21 forms 3,658*10^10 atoms of Rn per second.
Day has 246060=86400 s
That means that 2,66x10^21 atoms of Ra produces 3,16 x10^15 atoms of Rn in a day.
N(Rn)=3.16* 10 ^15 n(Rn)=N/NA
n(Rn)=5,25*10−9 pV=nR*T
T=273.15K R=8,314
p=101325Pa V=n∗R∗T/p
V=5.25∗10^−9 ∗ 8.314 ∗ 273.15 / 101325
V=1.1810^−10 m^3 = 118 x10^-7 liters of Rn, measured at STP, are produced per day by 1.0 g of Ra
To know more about Ra here
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Answer:
The answer to your question is: letter B
Explanation:
Reaction
Cr2O3(s) + 3CCl4(l) ⇒ 2CrCl3(s) + 3COCl2(g)
From the information given and the reaction, we can conclude that:
Green solid = Cr2O3 (s) "s" means solid
Colorless liquid = CCl4 (l) "l" means liquid and is the other reactant
Purple solid = CrCl3(s) CrCl3 is purple and "s" solid
Then, as a green specks remains it means that the excess reactant is Cr2O3, so, CCl4 is the limiting reactant.
Hope this helps. I provided step by step in the picture below if you want to see how I got these answers.
A= 1.0L
B= 0.50atm
C= 0.60atm
D= 4.0L
<h3>
Answer:</h3>
= 5.79 × 10^19 molecules
<h3>
Explanation:</h3>
The molar mass of the compound is 312 g/mol
Mass of the compound is 30.0 mg equivalent to 0.030 g (1 g = 1000 mg)
We are required to calculate the number of molecules present
We will use the following steps;
<h3>Step 1: Calculate the number of moles of the compound </h3>
Therefore;
Moles of the compound will be;
= 9.615 × 10⁻5 mole
<h3>Step 2: Calculate the number of molecules present </h3>
Using the Avogadro's constant, 6.022 × 10^23
1 mole of a compound contains 6.022 × 10^23 molecules
Therefore;
9.615 × 10⁻5 moles of the compound will have ;
= 9.615 × 10⁻5 moles × 6.022 × 10^23 molecules
= 5.79 × 10^19 molecules
Therefore the compound contains 5.79 × 10^19 molecules
