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Natali5045456 [20]
3 years ago
10

A metal pot is filled with water and placed on top of a stove. As the stove top heats up, the water begins to boil and steam can

be seen escaping from the pot. This is an example of a(n). Answer. A) closed system because matter can enter or leave the system but energy cannot.. B) open system because heat and matter can enter the system but cannot leave the system.. C) open system because heat and matter are both able to enter or leave the system.. D) closed system because energy can enter or leave the system but matter cannot..
Chemistry
2 answers:
Iteru [2.4K]3 years ago
5 0
A metal pot is filled with water and placed on top of a stove. As the stove top heats up, the water begins to boil and steam can be seen escaping from thepot.<span> </span>Open system. The matter can evaporate and leave in the form of steam or re condense and enter back into the system.
Darina [25.2K]3 years ago
3 0
Based on the scenario, the matter escaping from the pot is an example of an : C. Open system because heat and matter are both able to enter or leave the system

The key word here is 'steam can be seen escaping from the pot' , which indicate that the matter and heat able to leave the system after it's entered.

hope this helps
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In the three-dimensional structure of methane, ch4, the hydrogen atoms attached to a carbon atom are aligned ________. in the th
Nikitich [7]
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            In the three-dimensional structure of methane, CH₄, the hydrogen atoms attached to a carbon atom are aligned <span><u>at the corners of a Tetrahedron</u>.

Explanation:
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4 0
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An elemental analysis is performed in an unknown compound. It is found to contain 40.0 % mass in Carbon, 6.71% mass in Hydrogen,
DENIUS [597]

Answer: Molecular formula will be C_3H_6O_3

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 40 g

Mass of H = 6.71 g

Mass of O = 100 - (40+6.71 ) = 53.29 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40g}{12g/mole}=3.33moles

Moles of H =\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.71g}{1g/mole}=6.71moles

Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.29g}{16g/mole}=3.33moles

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

For C = \frac{3.33}{3.33}=1

For H =\frac{6.71}{3.33}=2

For O = \frac{3.33}{3.33}=1

The ratio of C: H: O= 1 : 2: 1

Hence the empirical formula is CH_2O

The empirical weight of CH_2O = 1(12)+2(1)+1(16)= 30g.

The molecular weight = 90.08 g/mole

Now we have to calculate the molecular formula:

n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{90.08}{30}=3

The molecular formula will be=3\times CH_2O=C_3H_6O_3

Molecular formula will be C_3H_6O_3

6 0
3 years ago
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