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Alex777 [14]
3 years ago
14

A sample of oxide iron weighing 2.40g was heated in a stream of hydrogen until it was completely converted to the metal. If the

metal weighed 1.68g. What is the formula of the iron oxide? (Fe=56, O=16)
Chemistry
1 answer:
wlad13 [49]3 years ago
4 0

Answer:

Fe₂O₃

Explanation:

To solve this question we must find the moles of Iron in 1.68g. With the difference of the masses we can find the moles of oxygen. The formula will be obtained with the ratio of both amount of moles:

<em>Moles Fe:</em>

1.68g * (1mol / 56g)  =0.03moles

<em>Moles O:</em>

2.40g-1.68g = 0.72g * (1mol/16g) = 0.045moles

The ratio O/Fe is:

0.045moles / 0.03moles = 1.5 moles. this ratio is obtained if the formula is:

<h3>Fe₂O₃</h3>
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How many miles of lead are equal to 9.51 x 10^3 g Pb?
Lemur [1.5K]
Molar mass Pb = 207.2 g/mol

1 mole Pb ------------- 207.2
? mole Pb ------------ 9.51 x 10³

moles = 9.51 x 10³ * 1 / 207.2

moles = 9.51 x 10³ / 207.2

= 45.89 moles

hope this helps!

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3 years ago
How do newtons 3 laws relate ​
Lina20 [59]

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3 years ago
Would nitrogen gas be a mineral if it has a chemical formula of n2?
STALIN [3.7K]

<u>Answer:</u>

Nitrogen gas be a mineral only, if it is in organic forms.

<u>Explanation:</u>

Most of the forms of organic nitrogen is not be taken by plants, with the exception in the form of small organic molecules. Also plants can promptly take the nitrogen  when it is in other forms like ammonia and nitrate.

The microorganisms  in the soil converts the organic forms of nitrogen to mineral form when they decompose organic matters and  also fresh plant residues. This type of process is called mineralisation.

6 0
3 years ago
Equal molar quantities of Ca2 and EDTA (H4Y) are added to make a 0.010 M solution of CaY2- at pH 10. The formation constant for
abruzzese [7]

Answer:

the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

Explanation:

Given the data in the question;

Ca^{2+ + y^{4- ⇄  CaY^{2-

Formation constant Kf

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) = 5.0 × 10¹⁰

Now,

[y^{4-] = \alpha _4CH_4Y; ∝₄ = 0.35

so the equilibrium is;

Ca^{2+ + H_4Y ⇄  CaY^{2- + 4H⁺

Given that; CH_4Y = Ca^{2+     { 1 mol Ca^{2+  reacts with 1 mol H_4Y  }

so at equilibrium, CH_4Y = Ca^{2+ = x

∴

Ca^{2+ + y^{4- ⇄  CaY^{2-

x        + x         0.010-x

since Kf is high, them x will be small so, 0.010-x is approximately 0.010

so;

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) =  CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ 0.010 / ( [x][ 0.35 × x] )  = 5.0 × 10¹⁰

⇒ 0.010 / 0.35x²  = 5.0 × 10¹⁰

⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 5.7142857 × 10⁻¹³

⇒ x = √(5.7142857 × 10⁻¹³)

⇒ x = 7.559 × 10⁻⁷

Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

8 0
2 years ago
A 0.885 M solution of KBr whose initial volume is 82.5 mL has more water added until its concentration is 0.500 M. What is the n
4vir4ik [10]

Answer:

V_2=146mL

Explanation:

Hello there!

In this case, since the equation for the calculation of dilutions is:

M_1V_1=M_2V_2

Whereas M is the molarity and V the volume, because the final concentration is lower than the initial. Thus, since we are asked to calculate the final volume, we solve for V2 as follows:

V_2=\frac{M_1V_1}{M_2}=\frac{0.885M*82.5mL}{0.500M}\\\\V_2=146mL

Best regards!

4 0
3 years ago
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