Answer:
317.6 mL
Explanation:
Step 1: Write the balanced neutralization equation
MgO + 2 HCl ⇒ MgCl₂ + H₂O
Step 2: Calculate the mass corresponding to 640.0 mg of MgO
The molar mass of MgO is 40.30 g/mol. The moles corresponding to 640.0 mg (0.6400 g) of MgO are:
0.6400 g × (1 mol/40.30 g) = 0.01588 mol
Step 3: Calculate the moles of HCl that react with 0.01588 moles of MgO
The molar ratio of MgO to HCl is 1:2. The moles of HCl are 2/1 × 0.01588 mol = 0.03176 mol
Step 4: Calculate the volume of 0.1000 M HCl that contains 0.03176 moles
0.03176 mol × (1 L/0.1000 mol) = 0.3176 L = 317.6 mL
Answer:
a) 1,6%
b) 64,775 g/mol
c) 3,6×10⁻² M
d) 2,3×10⁻³ g/mL
Explanation:
a) The mass fractium of helium is obtained converting the moles of the four gases to grams with molar weight and then caculating of the total of grams how many are of helium, thus:
- Helium: 0,25 moles ×
= 1 g of Helium - Argon: 0,25 moles ×
= 10 g of Argon - Krypton: 0,25 moles ×
= 20,95 g of krypton - Xenon: 0,25 moles ×
= 32,825 g of Xenon
Total grams: 1g+10g+20,85g+30,825g= 62,675 g
Mass fraction of helium: 
× 100 = <em>1,6%</em>
<em />
<em>The mass fraction of Helium is 1,6%</em>
<em />
<em>b)</em><em> </em>Because the mole fraction of all gases is the same the average molecular weight of the mixture is:
= 64,775 g/mol
c) The molar concentration is possible to know ussing ideal gas law, thus:
= M
Where:
P is pressure: 150 kPa
R is gas constant: 8,3145
T is temperature: 500 K
And M is molar concentration. Replacing:
M = 3,6×10⁻² M
d) The mass density is possible to know converting the moles of molarity to grams with average molecular weight and liters to mililiters, thus:
3,6×10⁻² 
× 
× 
=
2,3×10⁻³ g/mL
I hope it helps!
Ammonification is performed by bacteria to convert organic nitrogen to ammonia.
They allow us to see further and are able to focus on light from distant object. This is done by refracting or reflecting the light using lenses or mirrors.
Percent composition is the quantity or number of parts of a component contained in 100 parts of the total.
Given the the total is 100 g, the reported quantities for each component are the same percent.
Then, the percent composition of each as part of the entire sweet peas is:
Carb: 14.5%
Sugar: 5.7%
Diet. fiber: 5.1%
Prot: 5.4%
Fat: 0.4%
The rest to make the 100% is supposed to be water.
You could also find the percent composition in dry basis, which is done in this way (this is not what the question asks but it is good to know it).
Total components without water: 14.5 + 5.7 + 5.1 + 5.4 + 0.4 = 31.1 grams
Now divide each content by the total (31.1) and multiply by 100.
I will do one example:
Carbs: [14.5 / 31.1 ] * 100 = 46.6%
