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marshall27 [118]
3 years ago
14

Which element would chemically react with oxygen​

Chemistry
1 answer:
Makovka662 [10]3 years ago
5 0

Lithium, sodium, potassium, magnesium, calcium, barium, boron, aluminum, nitrogen, phosphorus, sulfur, fluorine and xenon are all examples of elements that react with oxygen. Oxygen reacts with other elements but those reactions are relatively mundane.

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Which shows the correct relationship between the molecular formula and the empirical formula for C2H10N2?
SIZIF [17.4K]
The correct answer is A. molecular formula = 2 x empirical formula

The given compound C2H10N2 is the molecular formula because the ratios of the elements are not reduced into lowest terms (empirical formula). To get the empirical formula, the ratios must be reduced to the smallest possible integral ratio which gives CH5N. As a result, the ratio between both formulas is 2.
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6b<br> Which of the following is a starting compound during<br> cellular respiration?
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Answer: Oxygen and glucose.

Explanation:

Oxygen and glucose are both reactants in the process of cellular respiration. The main product of cellular respiration is ATP; waste products include carbon dioxide and water.Jun 1, 2020

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I need help ASAP <br> Extra points &amp; brainlest:)
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Answer:

it is c

Explanation:

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8. What volume (ml) of a 5.45 M lead nitrate solution must be diluted to 820.7 ml to make a
Ganezh [65]

212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

Answer:

Option A.

Explanation:

Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

C_{1} V_{1} =C_{2} V_{2}

So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.

Then, (5.45*V_{1} ) = (820.7*1.41)

V_{1}=\frac{820.7*1.41}{5.45}=  212.33 ml

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

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As2O5
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