Answer:
0.200L of 0.300 Al(OH)₃ are necessaries
Explanation:
A diprotic acid, H₂X, reacts with aluminium (III) hydroxide, Al(OH)₃ as follows:
3 H₂X + 2 Al(OH)₃ → 6H₂O + Al₂X₃
To solve this question we must find the moles of the H₂X, using the balanced reaction we can find the moles of Al(OH)₃ and its volumen knowing its molar concentration is 0.300M:
<em>Moles H₂X:</em>
0.300L * (0.300mol / L) = 0.0900moles H₂X
<em>Moles Al(OH)₃:</em>
0.0900moles H₂X * (2mol Al(OH)₃ / 3mol H₂X) = 0.0600 moles Al(OH)₃
<em>Volume 0.300M Al(OH)₃:</em>
0.0600 moles Al(OH)₃ * (1L / 0.300moles) =
<h3>0.200L of 0.300 Al(OH)₃ are necessaries</h3>
Analyze the Name of complex Compound.
T<span>etracarbonylplatinum(iv) chloride
So, there are,
4 Carbonyl groups = 4 CO = (CO)</span>₄
1 Platinum Metal = 1 Pt = Pt
Unknown Chloride atoms = ?
In complexes positive part is always named first, so the sphere containing Pt and carbonyl ligands is written first,
[Pt (CO)₄]
The charge on sphere is +4 because CO ligand is neutral, and Pt has a Oxidation state of four as written in name (IV),
So,
[Pt (CO)₄]⁴⁺
Now, in order to neutralize +4 charge we should add 4 Chloride ions, So,
[Pt (CO)₄] Cl₄
<span>this is a limiting reagent problem.
first, balance the equation
4Na+ O2 ---> 2Na2O
use both the mass of Na and mass of O2 to figure out how much possible Na2O you could make.
start with Na and go to grams of Na2O
55.3 gNa x (1molNa/23.0gNa) x (2 molNa2O/4 molNa) x (62.0gNa2O/1molNa2O) = 75.5 gNa2O
do the same with O2
64.3 gO2 x (1 molO2/32.0gO2) x (2 molNa2O/1 mol O2) x (62.0gNa2O/1molNa2O) = 249.2 g Na2O
now you must pick the least amount of Na2O for the one that you actually get in the reaction. This is because you have to have both reacts still present for a reaction to occur. So after the Na runs out when it makes 75.5 gNa2O with O2, the reaction stops.
So, the mass of sodium oxide is
75.5 g</span>
Answer:C educated guess not completely sure
Explanation:
Answer: 110.2 lbs
Explanation: First, let's determine how many kg the patient masses. Since the dose is 3.000 mg/kg we can convert that to 0.003000 g/kg. So
0.1500 g / 0.003000 g/kg = 50 kg.
So we know the patient has a weight of 50 kg. Now we simply multiply by 2.20462 to get lbs. So
50 kg * 2.20462 lbs/kg = 110.231 lbs
Rounding to 1 decimal place gives 110.2 lbs
