Explanation:
•1:2:1 is the correct ratio of carbon to hydrogen to oxygen in glucose
Low pressure and high temperature will make real gasses behave like ideal gasses.
<span>decomposition of SrCO3 to SrO and CO2 =change in mass
moles of CO2 =(1.850 g - 1.445 g).
</span>Mass of <span>C<span>O2</span></span><span> in mixture: 1.850-1.445 = 0.405g
</span>0.405g/44.01 g/mol <span>C<span>O2</span></span><span> = 0.0092 moles </span><span>C<span>O2</span></span><span>.
</span>ratio of <span>C<span>O2</span></span><span> to SrO in Sr</span><span>C<span>O3</span></span><span> is 1:1
</span><span> mass ratio = 1.358/1.850 = 0.7341, </span>
or 73.41% Sr<span>C<span>O3</span></span><span>.
</span>hope this helps
Answer:
Mass percent of food dyes = 0.0616%
Explanation:
Given data:
Mass of candy = 47.9 g
Calories = 240
Mass of fat = 10 g
Mass of carbohydrate = 34 g
Mass of protein = 2 g
Mass of food dyes = 29.5 mg
Mass percent of food dyes = ?
Solution:
First of all we will convert the mg into g.
Mass of food dyes = 29.5 mg × 1g /1000 mg = 0.0295 g
Mass percent of food dyes = mass of food dyes / total mass× 100
Now we will put the values.
Mass percent of food dyes = 0.0295 g / 47.9 g × 100
Mass percent of food dyes = 0.000616 × 100
Mass percent of food dyes = 0.0616%
Answer & Explanation:
The molar mass of calcium chloride is 110.98 g/mol. We can use this information to solve this problem. We can set up our equation like this..
Multiply straight across on the top and straight across on the bottom.
Now divide.
So, there are 3.00 moles of calcium chloride contained in a 33 gram sample which is answer choice D.