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Aleonysh [2.5K]

2 years ago

10

A student has two compounds in two separate bottles but with no labels on either one. One is an unbranched alkane, octane, C8H18

; the other is 1-hexene, an unbranched alkene, C6H12. Based on your observations in this experiment, tell what you should see in the following tests: (a)- Water solubility (b)- Ligroin solubility (c)- Combustion (d)- Density versus water (e)- Bromine test (f)- Permanganate test

1 answer:

Aleks [24]2 years ago

7 0

Answer:

a) both substances are insoluble in water

b) both substances are soluble in ligroin

c) both substances suffer combustion, octane produces more CO₂ than hexene.

d) both substances are less dense than waterl, with hexene having the lowest density.

e) only hexene would react with bromine

f) only hexene would react with permanganate

Explanation:

a) both substances are non-polar and water is polar

b) both substances are non-polar and lingroin is non-polar

c) C₈H₁₈ + 17.5O₂ → 8CO₂ + 9H₂O

C₆H₁₂ + 9O₂ → 6CO₂ + 6H₂O

d) water = 997 kg/m³

ocatne = 703 kg/m³

hexene = 673 kg/m³

e) bromine test is used to detect unsaturations

f) permanganate test is used to detect unsaturations

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NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the te

makvit [3.9K]

Answer:

There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

Explanation:

To decrease the temperature of the solution there are necessaries:

4,184J/g°C×(5,0°C-25,0°C)×(100,0g+X) = -Y

8368J + 83,68J/gX = Y <em>(1)</em>

Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.

Also, the energy Y will be:

Y = 25700J/mol× $\frac{1mol}{80,043g}$ X

Y = 321J/g X <em>(2)</em>

Replacing (2) in (1)

8368J + 83,68J/g X = 321J/g X

8363J = 237,32J/gX

<em>X = 35,2g</em>

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Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

I hope it helps!

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2 years ago

The electron in a ground-state H atom absorbs a photon of wavelength 97.20 nm. To what energy level does it move?

Lunna [17]

Answer:

E = h v = 6.626 x $10^{-34}$ x 97.20 x $10^{-9}$ = 6.44 x $10^{-41}$ J = 4.01 x $10^{-22}$ eV

n = Energy Level = $-E^{0}$ / E = -13.6 / 4.01 x $10^{-22}$ = 3.39 x $10^{22}$ = 1.84 x $10^{11}$ energy level...

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2 years ago

On a clear day at sea level, with a temperature of 25 °C, the partial pressure of N2 in air is 0.78 atm and the concentration of

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Answer : The partial pressure of nitrogen gas is, 2.94 atm

Explanation:

According top the Henry's Law, the concentration of a gas in a liquid is directly proportional to the partial pressure of the gas.

$C\propto P$

$C=K_H\times P$

$K_H$ is Henry's constant.

or,

$\frac{C_1}{C_2}=\frac{P_1}{P_2}$

where,

$C_1$ = initial concentration of gas = $5.3\times 10^{-4}M$

$C_2$ = final concentration of gas = $2.0\times 10^{-3}M$

$P_1$ = initial partial pressure of gas = 0.78 atm

$P_2$ = final partial pressure of gas = ?

Now put all the given values in the above formula, we get the final partial pressure of the gas.

$\frac{5.3\times 10^{-4}M}{2.0\times 10^{-3}M}=\frac{0.78atm}{P_2}$

$P_2=2.94atm$

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2 years ago

Anyway, which is a spectator ion involved in the reaction of k2cro4(aq) and ba(no3)2(aq)?

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Rewrite the chemical equation as an ionic one; express all soluble salts- those with state symbol (aq)- as their constituting ions while leaving the insoluble (s) intact.

$2 \; \text{K}^{+}\; (aq)+ 2\; \text{CrO}_4^{-} \; (aq) + \text{Ba}^{2+} \; (aq) + 2 \; \text{NO}_3^{-} \; (aq) \\ \to \text{Ba}(\text{CrO}_4)_2 \; (s) + 2 \; \text{K}^{+} \; (aq) + 2 \; \text{NO}_3^{-} \; (aq)$

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