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Phoenix [80]
3 years ago
14

How many moles are in 150 g of Iron (III) oxide? (Fe203)

Chemistry
2 answers:
artcher [175]3 years ago
8 0

Answer:

The answer is 159.6882

ollegr [7]3 years ago
6 0

Answer:

0.939moles

Explanation:

no. of moles = mass/molar mass

= 150/159.7

= 0.939moles

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Answer: True

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5 0
3 years ago
Why does ice float after it crystallizes? Its mass is greater than water. Its density is greater than water. Its density is less
tekilochka [14]

Ice floats after it crystallizes because ITS DENSITY IS LESS THAN THAT OF WATER.

When a quantity of water is cools down by reducing its temperature, the molecules of the water lose kinetic energy and slow down in their movement. As the water is cooling down, it is volume is expanding. When the temperature reaches zero degree Celsius, the water becomes ice. At this point, the ice can float on water because its density is less than that of water; this is as a result of the spaces that now exist in the ice structure.

7 0
3 years ago
Read 2 more answers
A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base
tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

HN_3 + OH- ---> N^-_{3} + H_2O

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of HN_3 = \dfrac{0.001755}{0.0383} = 0.0458 M

Concentration of N^{-}_3 = \dfrac{ 0.001995 }{0.0383} = 0.0521 M

GIven that :

Ka = 1.9 x 10^{-5}

Thus; it's pKa = 4.72

pH =4.72 +  log(\dfrac{ \ 0.0521}{0.0458})

pH =4.72 + log(1.1376)

pH =4.72 + 0.05598

pH =4.77598

pH ≅ 4.80

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3 years ago
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Which atom would it be most difficult to remove an electron from
Norma-Jean [14]
I would be difficult to remove an electron from a Noble or Inert Gas (also known as the group 8 or 0 elements).  This is because they all have filled outermost shells and as such the outermost shell would be held tightly to the nucleus and as such make it difficult to remove.  Examples Helium, Neon, Argon, Xenon, Krypton and Radon 
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