The answer is 0.59 M.
Molar mass (Mr) of MgCl₂ is the sum of the molar masses of its elements.
So, from the periodic table:
Mr(Mg) = 24.3 g/l
Mr(Cl) = 35.45 g/l
Mr(MgCl₂) = Mr(Mg) + 2Mr(Cl) = 24.3 + 2 · 35.45 = 24.3 + 70.9 = 95.2 g/l
So, 1 mol has 95.2 g/l.
Our solution contains 55.8g in 1 l of solution, which is 55.8 g/l
Now, we need to make a proportion:
1 mole has 95.2 g/l, how much moles will have 55.8 g/l:
1 M : 95.2 g/l = x : 55.8 g/l
x = 1 M · 55.8 g/l ÷ 95.2 g/l ≈ 0.59 M
Answer:
I think the answer is the 3rd one
Answer:
False
Explanation:
Amplitude does not affect wavelength
Answer:
4.4×10² cm³
Explanation:
From the question given above, the following data were obtained:
Diameter (d) = 68.3 mm
Height (h) = 0.120 m
Volume (V) =?
Next, we shall convert the diameter (i.e 68.3 mm) to cm.
This can be obtained as follow:
10 mm = 1 cm
Therefore
68.3 mm = 68.3 mm / 10 mm × 1 cm
68.3 mm = 6.83 cm
Therefore, the diameter 68.3 mm is equivalent 6.83 cm.
Next, we shall convert the height (i.e 0.120 m) to cm. This can be obtained as follow:
1 m = 100 cm
Therefore,
0.120 m = 0.120 m/ 1 m × 100 cm
0.120 m = 12 cm
Therefore, the height 0.120 m is equivalent 12 cm.
Next, we shall determine the radius of the cylinder. This can be obtained as follow:
Radius (r) is simply half of a diameter i.e
Radius (r) = Diameter (d) /2
r = d/2
Diameter (d) = 6.83 cm
Radius (r) =?
r = d/2
r = 6.83/2
r = 3.415 cm
Finally, we shall determine the volume of the cylinder as follow:
Radius (r) = 3.415 cm
Height (h) = 12 cm
Volume (V) =?
Pi (π) = 3.14
V = πr²h
V = 3.14 × (3.415) ² × 12
V = 440 cm³
V = 4.4×10² cm³
Therefore, the volume of the cylinder is 4.4×10² cm³
Answer:
Your answer is C
Explanation:
When an acid and base react, the acidic hydrogen ion and the basic hydroxide ion in each acid and base neutralize each other and form water. Meanwhile the conjugate base and conjugate acid (the leftover compounds) react to form an ionic molecule, or a salt. (In chemistry, when an anion and a cation form an ionic bond the new molecule is called a salt). Hope this helps!
