Answer:
The molarity (M) of the following solutions are :
A. M = 0.88 M
B. M = 0.76 M
Explanation:
A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.
Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)
= 27 + 3(16 + 1)
= 27 + 3(17) = 27 + 51
= 78 g/mole
= 78 g/mole
Given mass= 19.2 g/mole
Moles = 0.246
Volume = 280 mL = 0.280 L
Molarity = 0.879 M
Molarity = 0.88 M
B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr
Molar mass of KBr = 119 g/mole
Given mass = 235.9 g
Moles = 1.98
Volume = 2.6 L
Molarity = 0.762 M
Molarity = 0.76 M
A melting point of over 700 C and a density of less than 2 g/cm3 can be observed for many group 2 elements. In this group, the density increases on moving down the group, whereas the melting point increases upto calcium and then starts decreasing.
Calcium, symbol Ca is the element with melting point around 840 C and density of 1.55 g/cm3 which is closest to the specified data range .
The given question is incomplete. the complete question is : If 5.15 g
is dissolved in enough water to make exactly 323 ml of solution, what is the molar concentartion of nitrate ion.
Answer: The molar concentartion of nitrate ion is 0.195.
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
Given : 5.15 g of is dissolved
Volume of solution = 323 ml
where,
n= moles of solute
= volume of solution = 323 ml
As 1 M of gives 3 M of 
ions.
Thus 0.065 M of gives = 
of ions.
The molar concentartion of nitrate ion is 0.195.
The General Equation of Alkane is as follow,
CnH₂n₊₂ ------- (1)
As m/z = 380
So,
CnH₂n₊₂ = 380
Or,
12n+ 1₂n₊₂ = 380
12n+ ₂n ₊ ₂ = 380
14n ₊ ₂ = 380
14n = 380 - 2
14n = 378
n = 378 / 14
n = 27
Now, putting value of n in eq. 1
C₂₇H₍₂₇ₓ₂₎₊₂
C₂₇H₅₆
Result:
n = C = 27
y = H = 56
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
