Answer:
a. 0.528 M ..
Explanation:
Hello!
In this case, since the given by-mass percent can be written as:
By using the density and molar mass of the solute, cadmium sulfate, we can compute the molarity, by also making sure we convert from mL to L of solution:
Thereby, the answer is a. 0.528 M .
Best regards.
<u>Answer:</u> The correct answer is Option 4.
<u>Explanation:</u>
Bromothymol blue, Bromocresol green and Thymol blue are the indicators which change their color according to the change in pH of the solution.
The pH range and color change of these indicators are:
- Bromothymol Blue: The pH range for this indicator is 6.0 to 7.5 and color change is from yellow to blue. It appears yellow below pH 6.0 and blue above pH 7.5
- Bromocresol green: The pH range for this indicator is 3.5 to 6.0 and color change is from yellow to blue. It appears yellow below pH 3.5 and blue above pH 6.0
- Thymol Blue: The pH range for this indicator is 8.0 to 9.6 and color change is from yellow to blue. It appears yellow below pH 8.0 and blue above pH 9.6
As, the highest pH of all the indicators is 9.6, so every indicator will appear blue above pH 9.6.
Hence, the correct answer is Option 4.
Hello!
We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100 g solution)
C: 83.7% = 83,7 g
H: 16.3% = 16.3 g
Let us use the above mentioned data (in g) and values will be converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:
We note that the values found above are not integers, so let's divide these values by the smallest of them, so that the proportion is not changed, let's see:
Note: So the ratio in the smallest whole numbers of carbon to hydrogen is 3:7, t<span>hus, the minimum or empirical formula found for the compound will be:
</span>
I hope this helps. =)
Answer:
Why do you need good toothpaste?
Explanation:
Answer:
Explanation:
Utilizing Rydber's equation:
ΔE = Z²Rh ( 1/n₁² - 1/n₂²) and substituting the values given ( using the Rydbers constant value in Joules ), we have
n=1 to n= infinity
ΔE = 3² x (1/1² - 0) x 2.18 x 10⁻^18 J = 2.0 x 10⁻¹⁷ J (1/infinity is zero)
n= 3 to n= infinity
ΔE = 3² x (1/3² - 0) x 2.18 x 10⁻^18 J = 2.28 x 10^-18 J
b. The wavelength of the emitted can be obtained again by using Rydberg's equation but this time use the constant value 1.097 x 10⁷ m⁻¹ given in the problem .
1/λ = Z²Rh (1/n₁² - 1/n₂²) 10 ⁻¹ = 3² x 1.097 x 10⁷ m⁻¹ x (1/1² - 1/3²) m⁻¹
1/λ =8.8 x 10⁷ m⁻¹ ⇒ λ =1.1 x 10^-8 m
λ = 1.1 x 10^-8 m x 1 x 10⁹ nm/m = 11 nm
