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anastassius [24]
3 years ago
11

Binary compound of oxygen and an unknown element,X, has the formula XO2 and is 69.55 mass % oxygen, what is the weight of elemen

t X?
Chemistry
1 answer:
marissa [1.9K]3 years ago
6 0

Answer:

Weight of element X = 14.01 gram

Explanation:

Mass percentage is calculated by using :

Percent =\frac{Mass\ of\ element}{Total\ mass}\times 100

O\ Percent=\frac{Mass\ of\ O}{Total\ mass\ of\ XO2}\times 100

Mass of O = 16.0 gram

let mass of X = X grams = ?

Mass of XO2 = mass of X + 2(mass of O)

Mass of XO2 = X + 2(16)

Mass of XO2 = X + 32

Total mass of O in the compound = 2(16) = 32 grams

Percent O = 69.55 %

Percent\ O =\frac{Mass\ of\ O}{Total\ mass\ of\ XO2}\times 100

69.55=\frac{32}{X+32}\times 100

\frac{69.55}{100}=\frac{32}{X+32}

0.6955=\frac{32}{X+32}

0.6955\times (X+32)=32

0.6955X+ 22.256=32

0.6955X=9.744

X=\frac{9.744}{0.6955}

X =14.01

weight of element X = 14.01 gram

14.01 is mass of nitrogen N

The correct formula of this element is NO2

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How much concentrated 18M sulfuric acid is needed to prepare 250mL of a 6.0M solution?
slavikrds [6]
C₁ * V₁ = C₂ * V₂ 

18 * V₁ = 6.0 * 250

18 V₁ = 1500

V₁ = 1500 / 18

V₁ = 83.33 mL

hope this helps!


5 0
3 years ago
you have been observing an insect that defends itself from enemies by secreting a caustic liquid. analysis of the liquid shows i
Alexus [3.1K]

pH of the buffer solution is 1.76.

Chemical dissociation of formic acid in the water:

HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq)

The solution of formic acid and formate ions is a buffer.

[HCOO⁻] = 0.015 M; equilibrium concentration of formate ions

[HCOOH] + [HCOO⁻] = 1.45 M; sum of concentration of formic acid and formate

[HCOOH] = 1.45 M - 0.015 M

[HCOOH] = 1.435 M; equilibrium concentration of formic acid

pKa = -logKa

pKa = -log 1.8×10⁻⁴ M

pKa = 3.74

Henderson–Hasselbalch equation: pH = pKa + log(cs/ck)

pH = 3.74 + log (0.015 M/1.435 M)

pH = 3.74 - 1.98

pH = 1.76

More about buffer: brainly.com/question/4177791

#SPJ4

4 0
1 year ago
A buffer is prepared by adding 12.0g of ammonium chloride (NH4Cl) to 250mL of 1.00 M NH3 solution?The pH is 9.3 Write the net io
solniwko [45]

Answer:- NH_3(aq)+H^+(aq)\rightleftharpoons NH_4^+(aq)

Explanations:- The solution we have is a buffer solution and we know that a buffer solution resists a change in its pH if a strong acid or base is added to it.

Here, the buffer solution we have is of a weak base and it's conjugate acid. So, a strong acid(nitric acid) is added to this buffer then it reacts with the base present in the buffer so that the acid could be neutralized. This is called buffer action.

The net ionic equation is written as:

NH_3(aq)+H^+(aq)\rightleftharpoons NH_4^+(aq)

Note that HNO_3 is a strong acid and nitrate ion is the spectator ion so it is not included in the net ionic equation.

8 0
3 years ago
How do atmospheric updrafts affect clouds?
Andru [333]

Answer:

Updrafts characterize a storm's early development, during which warm air rises to the level where condensation begins and precipitation starts to develop. In a mature storm, updrafts are present alongside downdrafts caused by cooling and by falling precipitation.

Hope it helps  

Have a great Day : P

6 0
3 years ago
How many mL of a stock solution of 2.00 M KNO3 are needed to prepare 100.0 mL of 0.15M KNO3? with work plz
Tatiana [17]
Using the law of dilution :

Mi x Vi =  Mf x Vf

2.00 x Vi = 0.15 x 100.0

2.00 x Vi = 15

Vi = 15 / 2.00

Vi = 7.5 mL

hope this helps!


3 0
3 years ago
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