Answer
I think it’s 5
Explanation
Answer:
good electrical conductor
lustre in appearance
Answer:
B.3/5p
Explanation:
For this question, we have to remember <u>"Dalton's Law of Partial Pressures"</u>. This law says that the pressure of the mixture would be equal to the sum of the partial pressure of each gas.
Additionally, we have a <em>proportional relationship between moles and pressure</em>. In other words, more moles indicate more pressure and vice-versa.

Where:
=Partial pressure
=Total pressure
=mole fraction
With this in mind, we can work with the moles of each compound if we want to analyze the pressure. With the molar mass of each compound we can calculate the moles:
<u>moles of hydrogen gas</u>
The molar mass of hydrogen gas (
) is 2 g/mol, so:

<u>moles of oxygen gas</u>
The molar mass of oxygen gas (
) is 32 g/mol, so:

Now, total moles are:
Total moles = 2 + 3 = 5
With this value, we can write the partial pressure expression for each gas:


So, the answer would be <u>3/5P</u>.
I hope it helps!
Don’t really understand what you’re asking but, if you’re asking how to read a graduated cylinder:
Look at the graduated cylinder at eye level, find the meniscus, whatever the meniscus is at is your answer.
Explanation:
The balanced equation of the reaction is given as;
Mg(OH)2 (s) + 2 HBr (aq) → MgBr2 (aq) + 2 H2O (l)
1. How many grams of MgBr2 will be produced from 18.3 grams of HBr?
From the reaction;
2 mol of HBr produces 1 mol of MgBr2
Converting to masses using;
Mass = Number of moles * Molar mass
Molar mass of HBr = 80.91 g/mol
Molar mass of MgBr2 = 184.113 g/mol
This means;
(2 * 80.91 = 161.82g) of HBr produces (1 * 184.113 = 184.113g) MgBr2
18.3g would produce x
161.82 = 184.113
18.3 = x
x = (184.113 * 18.3 ) / 161.82 = 20.8 g
2. How many moles of H2O will be produced from 18.3 grams of HBr?
Converting the mass to mol;
Number of moles = Mass / Molar mass = 18.3 / 80.91 = 0.226 mol
From the reaction;
2 mol of HBr produces 2 mol of H2O
0.226 mol would produce x
2 =2
0.226 = x
x = 0.226 * 2 / 2 = 0.226 mol
3. How many grams of Mg(OH)2 are needed to completely react with 18.3 grams of HBr?
From the reaction;
2 mol of HBr reacts with 1 mol of Mg(OH)2
18.3g of HBr = 0.226 mol
2 = 1
0.226 = x
x = 0.226 * 1 /2
x = 0.113 mol