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3 years ago

I will give you 50 pts

Chemistry

2 answers:

LekaFEV [45]3 years ago

8 0

An aquarium and Lake Erie can be their own ecosystem .

Option a and c

Explanation:

An ecosystem is a "community of living organisms" where there is an interaction between the non-living components of the environment and the living components i.e. living organisms, plants, animals, etc. acting as a system.

In the question above, An Aquarium will have its own ecosystem since it is an artificial ecosystem and the organisms present in it are selective and maintained in a restricted environment. While Lake Erie has its own natural ecosystem where all the Aquatic components are in conjunction with each other.

STatiana [176]3 years ago

5 0

Answer:

aquarium and Lake Erie and a dead tree

Explanation:

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Which polymer structure should be used if a strong and heat - resistant adhesive is required?

alexira [117]

Cross-linked polymers are strong and heat - resistant hence they can be used for this purpose.

A polymer is obtained by the combination of small molecules called monomers. A polymer consists of a regular repeating unit of small molecules called monomers.

A cross-linked polymer is a polymer in which covalent bonds are used to join polymer chains together. Cross-linked polymers are strong and heat - resistant hence they can be used for this purpose.

Learn more: brainly.com/question/17638582

3 0

3 years ago

A sample of gas contains 0.1700 mol of OF2(g) and 0.1700 mol of H2O(g) and occupies a volume of 19.5 L. The following reaction t

andreev551 [17]

Answer: The volume of the sample after the reaction takes place is 29.25 L.

Explanation:

The given reaction equation is as follows.

$OF_{2}(g) + H_{2}O(g) \rightarrow O_{2}(g) + 2HF(g)$

So, moles of product formed are calculated as follows.

$\frac{3}{2} \times 0.17 mol \\= 0.255 mol$

Hence, the given data is as follows.

$n_{1}$ = 0.17 mol, $n_{2}$ = 0.255 mol

$V_{1}$ = 19.5 L, $V_{2} = ?$

As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$

Substitute the values into above formula as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{19.5 L}{0.17 mol} = \frac{V_{2}}{0.255 mol}\\V_{2} = \frac{19.5 L \times 0.255 mol}{0.17 mol}\\= \frac{4.9725}{0.17} L\\= 29.25 L$

Thus, we can conclude that the volume of the sample after the reaction takes place is 29.25 L.

8 0

2 years ago

Glucose (C6H12O6) can be fermented to yield ethanol (CH3CH2OH) and carbon dioxide (CO2). C6H12O6⟶2CH3CH2OH+2CO2 The molar mass o

jekas [21]

Answer:

The % yield is 74.45 %

Explanation:

Step 1: The balanced equation

C6H12O6⟶2CH3CH2OH+2CO2

Step 2: Data given

Molar mass glucose = 180.15 g/mol

Molar mass of ethanol = 46.08 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Mass of glucose = 61.5 grams

Mass of ethanol = 23.4 grams

Step 3: Calculate moles of glucose

Moles glucose = Mass glucose / Molar mass of glucose

Moles glucose = 61.5 grams / 180.15 g/mol

Moles glucose = 0.341 moles

Step 4: Calculate moles of ethanol

1 mole of glucose consumed, produces 2 moles of ethanol and 2 moles of CO2

0.341 moles of glucose, will produce 2*0.341 = 0.682 moles of ethanol

Step 5: Calculate mass of ethanol

Mass ethanol = moles ethanol * Molar mass ethanol

Mass ethanol = 0.682 moles * 46.08 g/mol

Mass ethanol = 31.43 grams = theoretical mass

Step 6: Calculate % yield

% yield = actual mass / theoretical mass

% yield = (23.4 grams / 31.43 grams) * 100%

% yield = 74.45 %

The % yield is 74.45 %

7 0

3 years ago

In terms of bonds what would the molecule C6H12 be classified as

docker41 [41]

It has a double C=C bond so that means it's unsaturated, but it can also be a cyclic compound with only simple C-C bonds

5 0

3 years ago

Molar mass is the number of moles of a substance that have a mass of 1 gram.

Mekhanik [1.2K]

I think the answer is true.

3 0

3 years ago