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2 years ago

Flocabulary circulation

Chemistry

1 answer:

Anastasy [175]2 years ago

3 0

Answer: the system of blood, blood vessels and the heart that carries blood throughout the body

Explanation: i hope this helps

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Do all objects emit electromagnetic waves?

kherson [118]

Answer:

no, not all objects emit electromagnet waves.

Explanation: because lets say the object is carpet it has no magnetic waves in it so not all objects do.

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3 years ago

Compare the boiling points of 1-pentyne and 1-octyne.

Elan Coil [88]

1-pentyne consists of a carbon chain of 5 carbons one with a triple bond. 1-octyne is a carbon chain of 8 carbons with a triple bond at some point. It is known that the longer the carbon chain the higher the boiling point since more energy will be required to break the bonds between carbons. Based on this it is predicted that 1-octyne will have a higher boiling point than 1-pentyne.

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3 years ago

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Among the gases listed, which would have the fastest rate of effusion?

Delvig [45]

"CH4" is the one gas among the choices given in the question that <span>would have the fastest rate of effusion. The correct option among all the options that are given in the question is the second option or option "B". I hope that this is the answer that has actually come to your great help.</span>

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3 years ago

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Is syrup a compound or solution

S_A_V [24]

Answer:

Syrup is a mixture of thick sticky liquid consisting of a concentrated solution of sugar and water with or without the addition of a flavoring agent or medicinal substance.

Explanation:

so compound

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3 years ago

When a 3.00 grams sample of a compound containing only c, h, and o was completely burned, 1.17 grams of h2o and 2.87 grams of co

SVEN [57.7K]

Answer:- $CH_2O_2$

Solution:- From given masses of carbon dioxide and water we could calculate the moles that helps to calculate the moles of C and H.

Molar mass of carbon dioxide = 44 gram per mol

molar mass of water = 18.02 gram per mol

From given info, combustion of compound gives 1.17 grams of water and 2.87 grams of carbon dioxide. Let's calculate the moles of these:

$1.17gH_2O(\frac{1mol}{18.02g})$

= $0.0649molH_2O$

Similarly, $2.87gCO_2(\frac{1mol}{44g})$

= $0.0652molCO_2$

One mol of water has two moles of H. So, the moles of H would be two times the moles of water as calculated above.

So, moles of H = 2* 0.0649 = 0.1298 mol

One mol of carbon dioxide contains one mol of C. So, the moles of C would be equal to the moles of carbon dioxide calculated above.

moles of C = 0.0652 mol

Let's convert the moles of H and C to grams so that we could calculate the amount of oxygen present in the sample as:

grams of H in sample = 1.008 x 0.1298 = 0.1308 g

grams of C in sample = 12*0.0652 = 0.7824 g

If we subtract the sum of the masses of C and H from sample mass then it would give as the mass of oxygen since the sample has only C, H and O.

mass of O in sample = 3.00g - (0.1308 g + 0.7824 g)

= 3.00 g - 0.9132 g

= 2.0868 g

Let's convert these grams of oxygen to moles on dividing by it's atomic mass as:

$2.0868gO(\frac{1mol}{15.999g})$

= 0.130 mol O

Now, we have the moles of all the three atoms and we know that an empirical formula is the simplest whole number ratio of the moles of atoms. So, let's calculate the ratio. For this, we divide the moles of each by the least one of them.Looking at the moles, the least value is for carbon. So, let's divide the moles of each by the moles of C as:

C = $\frac{0.0652}{0.0652}$ = 1

H = $\frac{0.1298}{0.0652}$ = 2

O = $\frac{0.130}{0.0652}$ = 2

The ratio of C, H and O is 1:2:2. So, the simplest formula of the compound is $CH_2O_2$ .

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3 years ago