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kozerog [31]
3 years ago
6

What is the tendency of a solute to change the boiling point of a solid called?

Chemistry
2 answers:
Bumek [7]3 years ago
6 0
Boiling point elevation
Ludmilka [50]3 years ago
3 0
Boiling point elevation
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Science Tectonic Plates Cause and Effect (Converge, Diverge and Transform)
marusya05 [52]
When convergent happens:
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3 0
3 years ago
if 3.0 grams of aluminum and 6.0 grams of bromine react to form AlBr3, how many grams of product would theoretically be produced
Gelneren [198K]
1) Chemical reaction: 2Al + 3Br₂ → 2AlBr₃.
m(Al) = 3,0 g.
m(Br₂) = 6,0 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 3,0 g ÷ 27 g/mol.
n(Al) = 0,11 mol.
n(Br₂) = n(Br₂) ÷ m(Br₂).
n(Br₂) = 6 g ÷ 160 g/mol.
n(Br₂) = 0,0375 mol; limiting reagens.
n(Br₂) : n(AlBr₃) = 3 : 2.
n(AlBr₃) = 0,025 mol.
m(AlBr₃) = 0,025 mol · 266,7 g/mol.
m(AlBr₃) = 6,67 g.

2) m(Br₂) - all bromine reacts, so mass of bromine after reaction is zero grams (m(Br₂) = 0 g).
n(Al) = 0,11 mol - 0,025 mol = 0,085 mol.
m(Al) = 0,085 mol · 27 g/mol.
m(Al) = 2,295 g.
m(AlBr₃) = 6,67 g · 0,72 (yield of reaction).
m(AlBr₃) = 4,8 g.
n - amount of substance.
M - molar mass.

4 0
3 years ago
Consider the decomposition of red mercury(II) oxide under standard state conditions. )H0 T SFE ڮ( H M 0 H (a) Is the decompositi
sertanlavr [38]

The question is incomplete, complete question is :

Consider the decomposition of red mercury(II) oxide under standard state conditions.

2HgO(s)\rightarrow 2Hg(l)+O_2(g)

Given :

\Delta S_{HgO}^o=70.29 J/mol K

\Delta S_{Hg}^o=75.9 J/mol K

\Delta S_{O_2}^o=205.2 J/mol K

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol

(a) Is the decomposition spontaneous under standard state conditions?

(b) Above what temperature does the reaction become spontaneous?

Answer:

a) The decomposition is spontaneous under standard state conditions.

b)The reaction will spontaneous above -419.69 Kelvins.

Explanation:

2HgO(s)\rightarrow 2Hg(l)+O_2(g)

Given :

\Delta S_{HgO}^o=70.29 J/mol K

\Delta S_{Hg}^o=75.9 J/mol K

\Delta S_{O_2}^o=205.2 J/mol K

Entropy change of the reaction ; ΔS

\Delta S=[2\times \Delta S_{Hg}^o+1\times \Delta S_{O_2}^o]-[2\times \Delta S_{HgO}^o]

=[2\times 75.9 J/mol K+1\times 205.2 J/molK]-[2\times 70.29 J/molK]=216.42 J/mol K

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol  = -90830 J/mol K

1 kJ = 1000 J

At standard condition the value of temperature = T = 298 K

ΔG = ΔH - TΔS

ΔG = -90830 J/mol K - 298 K × 216.42 J/mol K = -155,323.16 J/mol

ΔG < 0 ( spontaneous)

The decomposition is spontaneous under standard state conditions.

b) Above what temperature does the reaction become spontaneous

Let the ΔG = 0

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol  = -90830 J/mol K

ΔG = ΔH - TΔS

0 = -90830 J/mol K - T × 216.42 J/mol K

T = -419.69 K

The reaction will spontaneous above -419.69 Kelvins.

6 0
3 years ago
Given 1kg=2.20 lbs, how many kilograms are in 125 lbs?<br> (Metric conversion)
Nadya [2.5K]

Answer:

56.82 Kg

Explanation:

Given data:

1 Kg = 2.20 lbs

Number of kilogram in 125 lbs = ?

Solution:

lbs is used for pound. lb is abbreviation of libra. It is Latin word meaning balance.

Both kilogram and pounds are units of mass. Pound is smaller unit than kilogram.

one Kg = 2.20 lbs

Number of kg in 125 lbs:

125 lbs × 1 Kg/2.20 lbs

125 lbs.Kg/2.20 lbs

56.82 Kg

7 0
4 years ago
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