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1 month ago

A sample of a compound contains 160 g of oxygen and 20.2 g of hydrogen. Give the compound's empirical formula.

2 answers:

4 0

The empirical formula of the compound can be derived as follows: Given Mass of O = 160 g, Mass of H = 20.2 g, Molar mass of O = 16 g/mole, Molar mass of H = 1 g/mole. Firstly, we convert the specified masses into moles. Moles of O are calculated, and Moles of H are calculated next. Then, we find the mole ratio by dividing each mole value by the smallest mole figure obtained. The ratio of O to H results in 1:2. The empirical formula reflects this ratio using subscripts. Thus, the empirical formula is established.

castortr0y [3K]1 month ago

3 0

Oxygen present in the compound = 160 g; Amount of oxygen in the compound = 20.2 g. Number of moles of oxygen = 160/16 = 10 moles. Number of moles of hydrogen = 20.2/1.01 = 20 moles. Therefore, the ratio of oxygen to hydrogen is 1:2. Thus, the empirical formula for the compound is H2O. I trust this answer assists you.

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A 50.0 mL Volumetric flask filled with aqueous solution of hydrogen peroxide has a total mass of 88.5 g. The mass of the volumet

alisha [2963]

Answer:

1.44 g/mL

Explanation:

The following information was derived from the provided question:

Volume (V) of H2O2 = 50 mL.

Total mass of flask and H2O2 = 88.5 g.

Mass of the flask = 16.5 g.

What is the density (D)?

Next, we will calculate the mass of hydrogen peroxide (H2O2).

This is determined as follows:

Total mass of flask and H2O2 = 88.5 g.

Mass of the flask = 16.5 g.

What is the mass of H2O2?

Mass of H2O2 = (Total mass of flask + H2O2) – (Mass of flask)

Mass of H2O2 = 88.5 – 16.5

Mass of H2O2 = 72 g

Finally, we will find the density of hydrogen peroxide (H2O2) as follows:

Volume (V) of H2O2 = 50 mL.

Mass (m) of H2O2 = 72 g.

What is the density (D)?

Density (D) = mass (m) divided by volume (V)

D = m/V

D = 72 g / 50 mL

D = 1.44 g/mL

In conclusion, the hydrogen peroxide (H2O2) density is 1.44 g/mL.

6 0

2 months ago

Which statement is true concerning the reaction N(g) + N(g) → N2(g) + energy?

alisha [2963]

Answer:

During this process, energy is released as a bond is formed.

Explanation:

The elements present on the left side of the reaction equation are known as reactants, while those on the right are called the products.

In this case, N(g) and N(g) are the reactants, and N2(g) along with energy are the products.

Post-reaction, N(g) has vanished and has been replaced by N2(g). It confirms that a bond exists between the two N(g).

Moreover, energy has been released as part of the products (indicated on the right side)

The accurate statement is ''A bond is formed and energy is released''

5 0

2 months ago

a 0.5678 of KHP required 26.64cm³ of NaOH to complete neutralization.calculate the molarity of the NaOH solution

lions [2927]

Answer:

Explanation:

0.5678 G X GRAMS

KHC8H4O4 + NaOH = NaKC8H4O4 + H2O

1 MOL 1 MOL

0.5678G X 204G/MOL = 0.00278 MOL KHC8H4O4

0.00278 MOL KHC8H4O4 X 1 MOLE NaOH/1 MOLE KHC8H4O4=0.00278 MOL NaOH

0.00278 MOL NaOH/26.26ml=0.106 molar

4 0

2 months ago

What volume of co2 gas at 645 torr and 800. k could be produced by the decomposition of 45.0 g of caco3? caco3(s) → cao(s) + co2

KiRa [2933]

In the reaction: <span>caco3(s) → cao(s) + co2(g), it is evident that
1 mol (which is 100 g) of CaCO3 yields 1 mol (which is 44 g) of CO2
Now, the molarity of CaCO3 present in the reaction system is
</span>= $\frac{weight of CaCO3 (g)}{gram molecular weight}$
= $\frac{45}{100}$ = 0.45 mol

Thus, 0.45 mol of CaCO3 leads to the formation of 0.45 mol of CO2.

According to the ideal gas equation, we have PV = nRT
V = $\frac{nRT}{P}$ .
Considering P = 645 torr = 0.8487 atm (because 1 atm = 760 torr)
In that case, V = $\frac{0.45 X 0.08206 X 800}{0.8487}$

= 34.8 l

5 0

1 month ago