Answer
is: 325.5 mL of the stock solution.
c₁(AgNO₃)
= 0.675 M.
<span>
V</span>₂(AgNO₃) = 0.879 L.<span>
c</span>₂(AgNO₃) = 0.250 M.<span>
V</span>₁(AgNO₃) = ?<span>
c</span>₁ -
original concentration of the solution, before it gets diluted.<span>
c</span>₂
- final concentration of the solution, after dilution.<span>
V</span>₁
- <span>volume to
be diluted.
V</span>₂ - <span>final volume after
dilution.
c</span>₁ · V₁ = c₂
· V₂.<span>
V</span>₁(AgNO₃) = c₂ · V₂ ÷ c₁.<span>
V</span>₁(AgNO₃) = 0.250 M · 0.879 L ÷ 0.675 M.<span>
V</span>₁(AgNO₃) = 0.325 L · 1000 mL = 325 mL.
<u>Answer:</u>
<em>5 molecules of oxygen will be present
</em>
<u>Explanation:</u>
<em>In this reaction combustion is taking place hence there must be presence of oxygen in it.
</em>
We don't know the initial amount of carbon molecules in hydrogen molecules present hence the equation goes:-
Mass of fuel 16.74 grams, Mass of carbon dioxide is 1198 gram and water is 0.0654 remaining mass of fuel is 16.70 g. Hence amount of fuel used up is 0.04 gram
<em>
</em>
<em>Weight of products is 0.1852 and weight of oxygen is 0.1452
</em>
<em>No of molecules will be
molecules
</em>
We are given the density of mercury which is equal to 13.6 g/ml. Substances which sink in this metal should have densities higher than the density of the metal, otherwise they would float. The substance thta would sink in mercury is C. gold
If the solution is BASIC than it will turn purple but if ACIDIC it will turn pink.
B, Because the classroom started with cylinder cups they should add 7mL into the cup and then transfer the 7mL on the cups