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3 years ago

A barometer can track the activity of storms at different altitudes. true of false

Chemistry

2 answers:

Bogdan [553]3 years ago

8 0

Its all the way round true 99.9% sure because they launch baometers up in the air and it reflects off of the satalght

Mazyrski [523]3 years ago

7 0

The answer is false i just took the quiz on edg.

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if you react 28 grams of butene with excess hydrogen how many grams of butane would you expect to get

liberstina [14]

Answer:

The correct answer is 29 grams.

Explanation:

Based on the given question, the reaction will be,

CH3CH=CHCH3 + H2 ⇒ CH3CH2CH2CH3

The molecular weight of butene is 56 grams per mole, and the molecular weight of butane is 58 grams per mole.

Thus, it can be said that 56 grams of butene reacts with hydrogen gas and produces 58 grams of butane.

Therefore, 28 grams of butene when reacts with hydrogen gas to give,

= 58/56 * 28 = 29 grams of butane.

Hence, the mass of butane produced will be 29 grams.

8 0

2 years ago

The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C

Vikentia [17]

Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

$ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})$

$ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})$

$1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})$

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

Normal boiling point ⇒ P=1 atm

Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

$ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})$

$ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})$

$1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})$

T₂=371.77 K= 98.62 °C

5 0

3 years ago

What reaction is depicted by the given equation: au3 3e− au

mrs_skeptik [129]

This is a reduction half-reaction. Initially, gold exists as $Au^{3+}$ , and after the reaction, it has been converted to elemental gold. To make this possible, three electrons must be added to $Au^{3+}$ . This fits within the definition of reduction reactions, where the reactant gains electrons in the reaction.

3 0

3 years ago

Which reactions are oxidation-reduction reactions? check all that apply. fes(s) 2hcl(aq) â†’ h2s(g) fecl2(g) agno3(aq) nacl(aq)

Dafna1 [17]

oxidation-reduction reactions are -

2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(g)
Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

For reaction,

2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(g)

On reactant side:

Oxidation state of Carbon = +2

Oxidation state of Oxygen = 0

On product side:

Oxidation state of Carbon = +4

Oxidation state of Oxygen = -2

Here, carbon's oxidation state is rising from +2 to +4. As a result, it is oxidizing and the oxygen's oxidation state is decreasing from 0 to -2. As a result, it is decreasing.

For reaction,

Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

When reacting:

Iron's oxidation state is +3.

Carbon's oxidation state is +2.

On product side:

Iron's oxidation state is zero.

Carbon's oxidation state is +4.

Here, carbon's oxidation state is rising from +2 to +4. As a result, it is being oxidized and the iron's oxidation state is changing from +3 to 0. As a result, it is decreasing.

To learn more about oxidation-reduction from given link

brainly.com/question/5794822

#SPJ4

7 0

1 year ago

Heat of vaporization is the amount of heat required to

deff fn [24]

I believe the answer would be c?? I’m not too sure though

5 0

2 years ago