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3 years ago

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A compound X has the following percentage composition 66.7% carbon, 11.1% hydrogen and 22.2% oxygen .Calculate the empirical for

mula of X.The relative molecular mass of X is 72 calculate the molecular formula

1 answer:

Digiron [165]3 years ago

4 0

Explanation:

c. h. o

66.7%. 11.1%. 22.2%

____. ____. ____

12. 1. 16

1.558. 11.1. 1.39. (divide by the smallest)

1. 8. 1

empirical formula=ch8o

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Must show units and how they cancelli 1.) 175 km to um 3.) 385 nm to dm 5.) 492 um tom 7.) 52 x 103 dm to mm 9.) 321x 1035 mm to

morpeh [17]

Explanation:

1.) 175 km to μm

$1 km=10^9 \mu m$

$175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m$

3.) 385 nm to dm

$1 nm=10^{-8} dm$

$385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm$

5.) 492 μm to m

1 μm = $10^{-6} m$

$492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m$

7.) $52\times 10^3$ dm to mm

1 dm = 100 mm

$52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm$

9.) $321\times 10^{35}$ mm to km

$1 mm = 10^{-6} km$

$321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km$

11.) $456\times 10^3$ m to km

m = 0.001 km

$456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km$

13.) $422\times 10^3$ m to nm

$1 m = 10^{9} nm$

$422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm$

15.) $4.87\times 10^{30}$ m to pm

$1 m = 10^{12} pm$

$4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm$

17.) $5.26\times 10^3$ m to um

1 m = $10^{6} \mu m$

$5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m$

19.) $1.25\times 10^{35}$ m to Mm

1 m = $10^{-6} Mm$

$1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm$

21.) $4.22\times 10^3$ Tm to nm

$1 Tm = 10^{21} nm$

$4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm$

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3 years ago

5) How many moles are in 5 Liters of H2 gas?

telo118 [61]

From 5 L to moles, just divide 5 by 22.4. I got 0.22 moles of H2.
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3 years ago

calculate the mass required to prepare 2.5 L of 1.0 M NaOH solution. Given that the molar mass for NaOH is 40 g/mol.

Helen [10]

Answer:

The required mass to prepare 2.5 L of 1.0 M NaOH solution is 100 g

Explanation:

We do this by preparing the equation:

Mass = concentration (mol/L) x volume (L) x Molar mass

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3 years ago

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