Explanation:
No. Isotopes are atoms of the same element with different atomic masses (due to the different number of neutrons)
For example, carbon exists as carbon-12 and carbon-14, which both have 6 protons but have 6 and 8 neutrons respectively.
Answer:
[CaSO₄] = 36.26×10⁻² mol/L
Explanation:
Molarity (M) → mol/L → moles of solute in 1L of solution
Let's convert the volume from mL to L
250 mL . 1L/1000 mL = 0.250L
We need to determine the moles of solute. (mass / molar mass)
12.34 g / 136.13 g/mol = 0.0906 mol
M → 0.0906 mol / 0.250L = 36.26×10⁻² mol/L
At STP volume is 22.4 L
Molar mass NO₂ = 46.0 g/mol
1 mole ---------- 22.4 L
? mole ---------- 11.4 L
moles = 11.4 * 1 / 22.4
moles = 11.4 / 22.4
= 0.5089 moles of NO₂
Mass NO₂ :
moles NO₂ * molar mass
0.5089 * 46.0
= 23.4094 g of NO₂
hope this helps!
Answer: The squirrel's balloon will be 0.86 L
Explanation:
To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.
The equation given by this law is:

where,
are initial pressure and volume.
are final pressure and volume.
We are given:

Putting values in above equation, we get:

Thus the squirrel's balloon will be 0.86 L
Answer:
pH = 4.543
Explanation:
- CH3CH2COOH + H2O ↔ CH3CH2COO- + H3O+
- pKa = - Log Ka
∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
∴ pKa = 4.87
⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
added 300 mL 0f 0.02 M NaOH:
⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)
⇒ <em>C</em> CH3CH2COOH = 0.048 M
⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M
mass balance:
⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)
charge balance:
⇒ [H3O+] + [Na+] = [CH3CH2COO-]
∴ [Na+] = 0.02 M
⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)
(2) in (1):
⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]
replacing in Ka:
⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])
⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]
⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]
⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0
⇒ [H3O+ ] = 2.867 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = 4.543