Answer:
In the third tube, the concentration is 0.16 ug/mL
Explanation:
In the first step, the solution is diluted by 5. Then, the concentration will be
20 ug/mL / 5 = 4 ug/mL
Then, in the second step this 4 ug / ml solution is diluted by a factor of five again:
4 ug /ml / 5 = 0.8 ug/mL
This solution is then diluted again by 5 and the concentration in the third tube will be then:
0.8 ug/mL / 5 = <u>0.16 ug/mL </u>
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Another way to calculate this is to divide the original concentration by the dilution factor ( 5 in this case) elevated to the number of dilutions. In this case:
Concentration in the third tube = 20 ug/mL / 5³ = 0.16 ug/mL
Answer:
0.0300 moles of H₂
Explanation:
The original equation is PV = nRT. We need to change this to show moles (n).
n = 
It's important to convert your values to match the constant (r) in terms of units.
30.0 kPa = 0.296 atm
2500 mL = 2.50 L
27 °C = 300 K
Now, plug those values in to solve:
n = 
- for the sake of keeping the problem clean, I didn't include the units but you should just to make sure everything cancels out :)
Finally, you are left with n = 0.0300 moles of H₂
Answer: This is due to a dominance of political parties over ... legislative process, the influence opposition has on amending the government bills and formal and effective
Explanation: hope u get it right
<h3>
Answer:</h3><h3>a) 9.033 × 10²³ particles</h3><h3>b) 4.068 × 10²⁴ particles</h3><h3>c) 1.51 × 10²³ particles</h3>
Explanation:
For us to answer these questions, we have to know two formulas:
- Number of particles = moles × Avogadro's Number
- Moles = Mass ÷ Molar Mass
Therefore:
a) particles of Na = 1.50 mol × (6.022 × 10²³) particles/mol
= 9.033 × 10²³ particles
b) particles of Pb = 6.755 mol × (6.022 × 10²³) particles/mol
= 4.068 × 10²⁴ particles
c) particles of Si
= (7.02 g ÷ 28.085 g/mol) × (6.022 × 10²³) particles/mol
= 1.51 × 10²³ particles
Sodium chloride , commonly known as salt (although sea salt also contains other chemical salts), is an ionic compound with the chemical formula NaCl, representing a 1:1 ratio of sodium and chloride ions. With molar masses of 22.99 and 35.45 g/mol respectively, 100 g of NaCl contains 39.34 g Na and 60.66 g Cl.
