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3 years ago

Fe(NO3)2 not sure how to get the oxidation numbers of all elements

Chemistry

1 answer:

SOVA2 [1]3 years ago

4 0

Fe(NO3)2=0
Fe+2(+3+3*-2)=0
Fe+2*-3=0
Fe=-(-6)
Fe=+6

So, Fe=+6, N=+3, O=-2

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What is an everyday example of covalent bonds and ionic bonds

ludmilkaskok [199]

Answer:

1. CARBON DIOXIDE- it is a covalent compound, which is used in soft/cold drinks and some other fluids as well , and use it in daily life.

2. HYDROGEN MONOXIDE- it is the normal or original or pure water which we drink everyday in our daily life and it is very important for our survival

Explanation:

3 0

3 years ago

The equilibrium constant for the reaction 2x(g)+y(g)=2z(g) is 2.25 . what would be the concentration of y at equilibrium with 2

Troyanec [42]

$[\text{Y}] \approx0.337\;\text{mol}\cdot\text{dm}^{-3}$ at equilibrium.

<h3>Explanation</h3>

Concentration for each of the species:

$[\text{X}] = \dfrac{n}{V} = 2\; \text{mol}\cdot \text{dm}^{-3}$ ;
$[\text{Y}] = \dfrac{n}{V} = 0\; \text{mol}\cdot \text{dm}^{-3}$ ;
$[\text{Z}] = \dfrac{n}{V} = 3\; \text{mol}\cdot \text{dm}^{-3}$ .

There was no Y to start with; its concentration could only have increased. Let the change in $[\text{Y}]$ be $+x \; \text{mol}\cdot \text{dm}^{-3}$ .

Make a $\textbf{RICE}$ table.

Two moles of X will be produced and two moles of Z consumed for every one mole of Y produced. As a result, the <em>change</em> in $[\text{X}]$ will be $+2\;x \; \text{mol}\cdot \text{dm}^{-3}$ and the <em>change</em> in $[\text{Z}]$ will be $-2\;x \; \text{mol}\cdot \text{dm}^{-3}$ .

$\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& & &&&\end{array}$ .

Add the value in the C row to the I row:

What's the equation of $K_c$ for this reaction? Raise the concentration of each species to its coefficient. Products go to the numerator and reactants are on the denominator.

$K_c = \dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]}$ .

$K_c = 2.25$ . As a result,

$\dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]} = \dfrac{(3-2x)^{2}}{(2+2x)^{2} \cdot x} = K_c = 2.25$ .

$(3-2\;x)^{2}= 2.25 \cdot(2+2\;x)^{2} \cdot x\\4\;x^{2} - 12 \;x + 9 = 2.25 \;(4\;x^{3} + 8 \;x^{2} + 4 \;x)\\4\;x^{2} - 12\;x + 9 = 9 \;x^{3} + 18\;x^{2} + 9\;x\\9\;x^{3} + 14\;x^{2} + 21\;x - 9 = 0$ .

The degree of this polynomial is three. Plot the equation $y = 9\;x^{3} + 14\;x^{2} + 21\;x - 9$ on a graph and look for any zeros. There's only one zero at $x \approx 0.337$ . All three concentrations end up greater than zero.

Hence the equilibrium concentration of Y: $0.337\;\text{mol}\cdot\text{dm}^{-3}$ .

7 0

3 years ago

The water in a 2500-L aquarium contains 1.25 g of copper. Calculate the concentration of copper in the water in ppm (remember 1

lorasvet [3.4K]

Answer:

0.5ppm

Explanation:

Step 1:

Data obtained from the question.

Volume of water = 2500L

Mas of Cu = 1.25 g

Step 2:

Determination of the concentration of Cu in g/L. This is illustrated below:

Volume of water = 2500L

Mas of Cu = 1.25 g

Conc. of Cu In g/L =?

Conc. g/L = Mass /volume

Conc. of Cu in g/L = 1.25/2500

Conc. of Cu in g/L = 5x10^–4 g/L

Step 3:

Conversion of the concentration of Cu in g/L to ppm. This is illustrated below

Recall:

1g/L = 1000mg/L

Therefore, 5x10^–4 g/L = 5x10^–4 x 1000 = 0.5mg/L

Now, we know that 1mg/L is equal to 1ppm.

Therefore, 0.5mg/L is equivalent to 0.5ppm

6 0

4 years ago

11.7 grams to kilograms

Goshia [24]

11.7 g = 0.0117 kg
........
:)

3 0

4 years ago

Among the following scenarios, which best demonstrates resilience when confronted with stress?

Valentin [98]

B. Everyone else aren't trying to grow, they only complain.

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3 years ago