Answer is 2m/s sq because V-u /t us acceleration 40-10/15=2
Answer:
5.645 × 10⁻²³ g
Solution:
Step 1) Calculate Molar Mass of SH₂;
Atomic Mass of Sulfur = 32 g/mol
Atomic Mass of H₂ = 2 g/mol
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Molecular Mass of SH₂ = 34 g/mol
Step 2: Calculate mass of one molecule of SH₂ as;
As,
Moles = # of Molecules / 6.022 × 10²³
Also, Moles = Mass / M.Mass So,
Mass/M.mass = # of Molecules / 6.022 × 10²³
Solving for Mass,
Mass = # of Molecules × M.mass / 6.022 × 10²³
Putting values,
Mass = (1 Molecule × 34 g.mol⁻¹) ÷ 6.022 × 10²³
Mass = 5.645 × 10⁻²³ g
TITRATION is the process of reaching equilibrium between acids and bases.
Explanation:
here's the answer to your question
Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
