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2 years ago

How many grams of CO2 are produced from 51 g of CH4

Chemistry

1 answer:

Elina [12.6K]2 years ago

3 0

Answer:

use the equation Mass= RFM*Moles

Explanation:

use your periodic table

and create a little table

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Combustion analysis of 0.300 g of an unknown compound containing carbon, hydrogen, and oxygen produced 0.5213 g of co2 and 0.283

Lyrx [107]

First, we have to get how many grams of C & H & O in the compound:
- the mass of C on CO2 = mass of CO2*molar mass of C /molar mass of CO2
= 0.5213 * 12 / 44 = 0.142 g
- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
=0.2835 * 2 / 18 = 0.0315 g
- the mass of O = the total mass - the mass of C atom - the mass of H atom
= 0.3 - 0.142 - 0.0315 = 0.1265 g
Convert the mass to mole by divided by molar mass
C(0.142/12) H(0.0315/2) O(0.1265/16)
C(0.0118) H(0.01575) O(0.0079) by dividing by the smallest value 0.0079
C1.504 H3.99 O1 by rounding to the nearst fraction
C3/2 H4/1 )1/1 multiply by 2
∴ the emprical formula C3H8O2

6 0

2 years ago

Cl2+2KI→2KCl + I2 is an example of which type of reaction?

Gnoma [55]

Answer:

single replacement

Explanation:

3 0

2 years ago

a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding empirical formula is

fiasKO [112]

Answer: The empirical formula is $CH_2$ .

Explanation:

Mass of C = 1.71 g

Mass of H = 0.287 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{1.71g}{12g/mole}=0.142moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.287g}{1g/mole}=0.287moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.142}{0.142}=1$

For H = $\frac{0.287}{0.142}=2$

The ratio of C: H = 1: 2

Hence the empirical formula is $CH_2$ .

4 0

2 years ago

Question 1. In the following equation for the alpha decay of bismuth-214, supply the missing element that correctly completes th

Klio2033 [76]

Q1. TI (210/81Thallium)
Q2.
The answers are opposite from each other

7 0

2 years ago

Read 2 more answers

Propanoic acid has a boiling point of 141 °C, propanamide has a boiling point of 213 °C, and propanal has a boiling point of 48

deff fn [24]

Explanation:

As it is given that boiling point of propanamide is very high. So, reason for this is that easy formation of hydrogen bonds which are strong enough that we have to provide large amount of heat to break it.

As in $-NH_{2}$ , the hydrogen atoms which are present are positive in nature. Due to this they are able to form hydrogen bonds with the neighboring oxygen atom.

Hence, these bonds are so strong that high heat needs to given to break them.

A propanoic acid contain carboxylic group as the functional group. So, this group is also able to form hydrogen bonding as it forms a hydrogen bond between an acid group and hydroxyl group of neighboring molecule.

Hence, it will also require high heat to break the bond due to which there will be increase in boiling point.

In propanal, there is presence of aldehyde functional group and three carbon atoms chain which will not form strong bonding with the hydrogen atom of CHO. Due to this there will exist weak Vander waal's force that is not at all strong enough.

As a result, less energy will be needed to break the bonds in propanal. Hence, it has very low boiling point.

4 0

2 years ago