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3 years ago

QUE ES UN ENLACE GLUCOSÌDICO?

Chemistry

1 answer:

Blizzard [7]3 years ago

8 0

Answer:

Explanation:

no se hombre necisito pontos

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What produces carbon dioxide?

Andrews [41]

Burning fossil fuels

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3 years ago

Read 2 more answers

A 0.211 g sample of carbon dioxide, CO2, has a volume of 560 mL and a pressure of 429 mmHg whats the temperature of the gas in K

SCORPION-xisa [38]

The temperature of the gas sample is 813 K.

Explanation:

We have to use the ideal gas equation to find the temperature of the gas sample.

The ideal gas equation is PV = nRT

Pressure, P = 429 mm Hg = 0.56 atm

Volume, V = 560 mL = 0.56 L

R = gas constant = 0.08205 L atm mol⁻¹K⁻¹

Mass = 0.211 g

Molar mass of carbon di oxide = 44.01 g / mol

Moles, n = $$\frac{given mass}{molar mass} = \frac{0.211 g}{44.01 g/mol}$

= 0.0047 mol

Now, we have to plugin the above values in the above equation, we will get the temperature as,

$$T= \frac{PV}{nR}$

T = $$\frac{0.56 \times 0.56}{0.08205 \times 0.0047}$

= 813 K

So the temperature of the gas sample is 813 K.

5 0

4 years ago

you have two solid substances that look the same. what measurements would you take and what tests would you perform to determine

andreev551 [17]

The easiest way to go about this would be to check out their physical properties. Physical properties are those attributes of a substance or material that can be observed or measured without altering or changing the composition of that material.

So we can check out the following physical properties:

1. Color. Check if there is any difference in color.

2. Weight. If possible, take similar sized pieces of the two and note any difference in weight .

3. Texture. Are they smooth or rough to the touch?

4. Smell or odor. Check out any unique smell.

5. Solubility. Dissolve a bit of each in water and see how soluble each is.

6. Melting point. Carefully heat a piece of it and see if they will melt, and if so at what temperature.

7. Pliability. Can any of them bend, are they flexible?

8. Magnetism. Can the be attracted by a magnet?

9. Bouyancy . Can they float on water?

10. Electrical conductivity. using a battery and small torch bulb, find out if they can conduct an electric current.

These 10 physical tests can help differentiate between the two materials. There are still many more that can be safely carried out.

8 0

3 years ago

How does density of air related to elevation (altitude)

dem82 [27]

Answer:

The density of air goes down as you get higher with altitude.

Explanation:

Less air is pushing down on the molecules below it as you go up. This means that it is not as compacted, another word for which is dense.

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3 years ago

Ethylene (CH2CH2) is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about 8.0 x 10

mylen [45]

Answer: The percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

Explanation:

We are given:

Initial partial pressure or ethane = 24.0 atm

The chemical equation for the dehydration of ethane follows:

$C_2H_6(g)\rightleftharpoons C_2H_4(g)+H_2(g)$

Initial: 24.0

At eqllm: 24-x x x

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{C_2H_4}\times p_{H_2}}{p_{C_2H_6}}$

We are given:

$K_p=0.040$

Putting values in above expression, we get:

$0.040=\frac{x\times x}{24-x}\\\\x^2+0.04x-0.96=0\\\\x=0.96,-1$

Neglecting the value of x = -1 because partial pressure cannot be negative.

So, partial pressure of hydrogen gas at equilibrium = x = 0.96 atm

Partial pressure of ethylene gas at equilibrium = x = 0.96 atm

Partial pressure of ethane gas at equilibrium = (24-x) = (24 - 0.96) atm = 23.04 atm

To calculate the number of moles, we use the equation given by ideal gas, which follows:

$PV=nRT$ .........(1)

To calculate the mass of a substance, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ..........(2)

For ethane gas:

We are given:

$P=23.04atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$23.04atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{23.04\times 30.0}{0.0821\times 1073}=7.85mol$

We know that:

Molar mass of ethane gas = 30 g/mol

Putting values in equation 2, we get:

$7.85mol=\frac{\text{Mass of ethane gas}}{30g/mol}\\\\\text{Mass of ethane gas}=(7.85mol\times 30g/mol)=235.5g$

For ethylene gas:

We are given:

$P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol$

We know that:

Molar mass of ethylene gas = 28 g/mol

Putting values in equation 2, we get:

$0.33mol=\frac{\text{Mass of ethylene gas}}{28g/mol}\\\\\text{Mass of ethylene gas}=(0.33mol\times 28g/mol)=9.24g$

For hydrogen gas:

We are given:

$P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol$

We know that:

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 2, we get:

$0.33mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.33mol\times 2g/mol)=0.66g$

To calculate the mass percentage of ethylene in equilibrium gas mixture, we use the equation:

$\text{Mass percent of ethylene gas}=\frac{\text{Mass of ethylene gas}}{\text{Mass of equilibrium gas mixture}}\times 100$

Mass of equilibrium gas mixture = [235.5 + 9.24 + 0.66] = 245.4 g

Mass of ethylene gas = 9.24 g

Putting values in above equation, we get:

$\text{Mass percent of ethylene gas}=\frac{9.24g}{245.5g}\times 100=3.76\%$

Hence, the percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

5 0

4 years ago