Answer:

Explanation:
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In this case, since the normal boiling point of X is 117.80 °C, the boiling point elevation constant is 1.48 °C*kg*mol⁻¹, the mass of X is 100 g and the boiling point of the mixture of X and KBr boils at 119.3 °C, we can use the following formula:

Whereas the Van't Hoff factor of KBr is 2 as it dissociates into potassium cations and bromide ions; it means that we can compute the molality of the solution:

Next, given the mass of solventin kg (0.1 kg from 100 g), we compute the moles KBr:

Finally, considering the molar mass of KBr (119 g/mol) we compute the mass that was dissolved:

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Answer:
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Explanation:
The reaction:
2 H 2 + O 2 → 2 H 2 O
2 Hydrogen molecules react with 1 Oxygen molecule to create 2 molecules of water.
We have to convert 53.8 g of Hydrogen into moles:
53.8 : 2.02 g / moles = 26.63 moles
2 moles H 2 ↔ 1 mole O 2
26.63 moles H 2 ↔ 13.32 moles O 2
Mass ( O 2 ) = 13.32 moles · 32 g/moles = 426.2 g
Answer: 426.2 grams of Oxygen.
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
Answer:
Mass of 1 litre of this gas at 273 degree Celsius and 1140 mm Hg pressure is
3 grams. So, (T1) = (273 + 273) degree Absolute = 546 degree Absolute and
(P1) = (760 + 1140) mm Hg = 1900 mm Hg and (V1) = 1 Litre
First, we have to find out how much volume does it occupy at NTP; (T2) = 273 degree Absolute and (P2) = 760 mm Hg.
So, (V2) = (1900*1 / 546)*(273 / 760) = 1.25 litre.
So, the mass of 1.25 litre volume of this gas = 3 grams
Therefore, the mass of 22.4 litre volume of this gas = (3*22.4 / 1.25) grams
= 53.76 grams.
So, the gram molecular mass of the gas is 53.76.
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