Answer:
Here's what I find.
Explanation:
Iodine-131
Iodine-131 is both a beta emitter and a gamma emitter.
About 90 % of the energy is β-radiation and 10 % is γ-radiation. Both forms are highly energetic.
The main danger is from ingestion. The iodine concentrates in thyroid gland, where the β-radiation destroys cells up to 2 mm from the tissues that absorbed it.
Both the β- and γ-radiation cause cell mutations that can later become cancerous. Small doses, such as those absorbed from the nuclear disasters in the Ukraine and Japan, can cause cancers years after the original iodine has disappeared.
Plutonium-239
Plutonium-239 is an alpha emitter.
Alpha particles cannot penetrate the skin, so external exposure isn't much of a health risk.
However, they are extremely dangerous when they are inhaled and get inside cells. They travel first to the blood or lymph system and later to the bone marrow and liver, where they cause up to 1000 times more chromosomal damage than beta or gamma rays.
It takes about 20 years for plutonium to be eliminated from the liver around 50 years for from the skeleton, so it has a long time to cause damage.
Answer:
Determine the frequency of light with a wavelength of 2.775⋅10−7 cm. Answer in units of Hz?❤
Answer:try to make a good pic please I have ideas about the topic
Answer:
1. V₁ = 2.0 mL
2. V₁ = 2.5 mL
Explanation:
<em>You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.</em>
To calculate the volume required (V₁) in each dilution we will use the dilution rule.
C₁ . V₁ = C₂ . V₂
where,
C are the concentrations
V are the volumes
1 refers to the initial state
2 refers to the final state
<em>1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.</em>
C₁ . V₁ = C₂ . V₂
(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL
V₁ = 2.0 mL
<em>2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.</em>
C₁ . V₁ = C₂ . V₂
(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL
V₁ = 2.5 mL
Answer:
25.35%
Explanation:
Again let me restate the the equation of the reaction;
H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)
Amount of potassium permanganate reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles
If 2 moles of MnO4 - reacts with 3 moles of CN-
8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2
= 1.229 * 10^-3 moles of CN-
Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol
= 0.03 g
Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100
= 25.35%
