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n200080 [17]
3 years ago
12

Calculate the mass of Iron in 40g of Fe203​

Chemistry
1 answer:
Nat2105 [25]3 years ago
7 0

Answer:

   ‏‏‎ ‎

Explanation:‏‏‎ ‎‏‏‎ ‎‏‏‎ ‎

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A student makes an aqueous solution of sodium hydroxide. Which statement correctly identifies the two ions present in the solute
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Answer:

A. Na+ OH-

Explanation:

positive and negative ions form bonds because positive and negative attract

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2 years ago
Some nickel sulfate solution was carefully placed in the bottom of a beaker of water. The beaker was then covered and left for s
Furkat [3]
When’s this due I think I can find out
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1) Assume element X has 2 isotopes: X-125 and X-126. For every 100 atoms of X, 30 of them have a mass of 125.0 u and 70 have a m
harkovskaia [24]

Answer:

Explanation:

 .3(125)+.7(126)=37.5+88.2=125.7

2. The most common isotope is 32, because the average is very close to 32.

6 0
3 years ago
Read 2 more answers
Q1. Calculate the amount of copper produced in 1.0 hour when aqueous CuBr2 solution was electrolyzed by using a current of 4.50
OLga [1]

Answer:

\boxed{\text{Q1. 3.6 g; Q2. 0.2 A}}

Explanation:

Q1. Mass of Cu

(a) Write the equation for the half-reaction.

Cu²⁺ + 2e⁻ ⟶ Cu

The number of electrons transferred (z) is 2 mol per mole of Cu.

(b) Calculate the number of coulombs

q  = It  

\text{t} = \text{1.0 h} \times \dfrac{\text{3600 s}}{\text{1 h}} = \text{3600 s}\\\\q = \text{3 C/s} \times \text{ 3600 s} = \textbf{10 800 C}

(c) Mass of Cu

We can summarize Faraday's laws of electrolysis as

\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\& = &\dfrac{10 800 \times 63.55}{2 \times 96 485}\\\\& = & \textbf{3.6 g}\\\end{array}\\\text{The mass of Cu produced is $\boxed{\textbf{3.6 g}}$}

Note: The answer can have only two significant figures because that is all you gave for the time.

Q2. Current used

(a) Write the equation for the half-reaction.

Ag⁺ + e⁻ ⟶ Ag

The number of electrons transferred (z) is 1 mol per mole of Ag.  

(a) Calculate q

\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\2.00& = &\dfrac{q \times 107.87}{1 \times 96 485}\\\\q &=& \dfrac{2.00 \times 96485}{107.87}\\\\& = & \textbf{1789 C}\\\end{array}

(b) Calculate the current

t = 3 h = 3 × 3600 s = 10 800 s

\begin{array}{rcl}q&=& It\\1789 & = & I \times 10800\\I & = & \dfrac{1789}{10800}\\\\& = & \textbf{0.2 A}\\\end{array}\\\text{The current used was $\large \boxed{\textbf{0.2 A}}$}

Note: The answer can have only one significant figure because that is all you gave for the time.

4 0
3 years ago
Which of the mole ratios would be used to set up the below problem?
san4es73 [151]
A.3 moles H2

1 mole N2

I think it is correct
4 0
3 years ago
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