94.6 g. You must use 94.6 g of 92.5 % H_2SO_4 to make 250 g of 35.0 % H_2SO_4.
We can use a version of the <em>dilution formula</em>
<em>m</em>_1<em>C</em>_1 = <em>m</em>_2<em>C</em>_2
where
<em>m</em> represents the mass and
<em>C</em> represents the percent concentrations
We can rearrange the formula to get
<em>m</em>_2= <em>m</em>_1 × (<em>C</em>_1/<em>C</em>_2)
<em>m</em>_1 = 250 g; <em>C</em>_1 = 35.0 %
<em>m</em>_2 = ?; _____<em>C</em>_2 = 92.5 %
∴ <em>m</em>_2 = 250 g × (35.0 %/92.5 %) = 94.6 g
The answer is the troposphere
Answer:
i) CCl₄ and Br₂ does not react
ii) CBr₄ + Cl₂ → CCl₄ + Br₂
Explanation:
i) CCl₄ + Br₂ (no reaction)
From the given activity series, we have that chlorine gas, Cl₂, is more reactive than bromine gas, Br₂, therefore, a reaction of CCl₄ + Br₂ will not have a reaction as the propensity for the chlorine to stay combined with the carbon is higher than the ability for bromine to remain combined with or attract the carbon. Therefore, for CCl₄ + Br₂ there is no reaction
ii) CBr₄ + Cl₂
From the given activity series, we have that chlorine gas, Cl₂, is more reactive than bromine gas, Br₂, therefore, a reaction of CBr₄ + Cl₂ will give products that will have the Br in the CBr₄ replaced by the Cl₂ as follows;
CBr₄ + Cl₂ → CCl₄ + Br₂
The products of the reaction of CBr₄ and Cl₂ are therefore CBr₄ and Cl₂.
1. Convert the number of moles to number of particles (divide 6.41 by 6.022 x 10^23)
2. K2O dissociates into three ions (2 potassium ions and 1 oxide ion), so multiply the result from step 1 by 3 to get your answer
