Answer:
b 13kg/m³ is the answer to this question
Half-life is defined as the amount of time it takes a given quantity to decrease to half of its initial value. The equation to describe the decay is
Nt=N0(1/2)
where N0 is the initial quantity, Nt is the remaining quantity after time t, t1/2 is the half-time. So work out the equation, t1/2 = t (-ln2)/ln(Nt/N0) = 11.5*(-ln2)/ln(12.5/100) = 3.83 days
An ionization suppressor is an alkali metal capable of preventing ionization, which can be used in atomic spectroscopy to determine matter composition.
<h3>What is ionization?</h3>
Ionization refers to the phenomena capable of converting neutral atoms/molecules to electrically charged atoms/ions.
Ionization is a process by which radiation (e.g., alpha, beta, gamma rays) can pass energy to inert matter.
Some examples of ionization suppressors include salts of alkali metals (for example, potassium), which can be used in atomic spectroscopy to determine matter composition.
Learn more about ionization here:
brainly.com/question/1445179
Answer:
<u>C) 4</u>
Explanation:
<u>The reaction</u> :
- C (s) + 2H₂ (g) ⇒ CH₄ (g)
12g 4g 16g
Hence, based on this we can say that : <u>2 moles of hydrogen gas are needed to produce 16g of methane.</u>
<u />
<u>For 32g of methane</u>
- Number of moles of H₂ = 32/16 × 2
- Number of moles of H₂ = <u>4</u>
Answer:
The mass in grams of glucose produced when 132.0 g of CO2 reacts with an excess of water is 90.1 grams
Explanation:
The chemical equation for the reaction is
6H₂O + 6CO₂ → C₆H₁₂O₆ + 6O₂
From the reaction, it is seen that 6 moles of H₂O reacts ith 6 moles of CO₂ to produce 1 mole of glucose C₆H₁₂O₆ and 6 moles oxygen gas
The molar mass of CO₂ = 44.01 g/mol
There fpre 132.0 g contains 132.0/44.01 moles or ≅ 3 moles
However since 6 moles of CO₂ produces 1 mole of O₂, then 3 moles of CO₂ will prduce 1/6×3 or 0.5 moles of C₆H₁₂O₆
and since the molar mass (or the mass of one mole) of C₆H₁₂O₆ is 180.2 grams/mole then 0.5 mole of C₆H₁₂O₆ will have a mass of
mass of 1 mole C₆H₁₂O₆ = 180.2 g
mass of 0.5 mole C₆H₁₂O₆ = 180.2 g × 0.5 = 90.1 grams
Mass of glucose produced = 90.1 grams