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4 years ago

Which element would not have similar properties with Calcium?

Chemistry

2 answers:

MakcuM [25]4 years ago

7 0

K potassium would be the answer

elena-14-01-66 [18.8K]4 years ago

5 0

Potassium would not have similar properties because it is the alkali metals, not alkali earth metals

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How many grams of NaOH would be required to make 1.0 L of a 1.5 M solution

omeli [17]

60 grams are required.

Hope this helped you!

4 0

3 years ago

All of the following statements concerning crystal field theory are true EXCEPT Group of answer choices in an isolated atom or i

yaroslaw [1]

Answer:

low-spin complexes contain the maximum number of unpaired electrons.

Explanation:

In the crystal field theory, the magnitude of crystal field splitting and the pairing energy determines whether a complex will be low spin or high spin.

Low spin complexes often have greater magnitude of crystal field splitting energy than low spin complexes.

High spin complexes have maximum number of unpaired electrons(most of the electrons are unpaired) while low spin complexes have a minimum number of unpaired electrons in a complex(most of the electrons are paired).

6 0

3 years ago

Suppose you are given an unknown solid material. Describe some tests you might perform to prove that is either an element or a c

nikitadnepr [17]

<span>An element is a pure substance that cannot be broken down by further chemical techniques. Therefore you could try heating, cooling, electrolysis or other methods to see if it was an element or a compound.</span>

5 0

3 years ago

Write the equilibrium constant expression for this reaction: 2H+(aq)+CO−23(aq) → H2CO3(aq)

MrRissso [65]

Answer:

Equilibrium constant expression for $\rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq) \rightleftharpoons H_2CO_3\, (aq)$ :

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{[\mathrm{H_2CO_3}]}{\left[\mathrm{H^{+}\, (aq)}\right]^{2} \, \left[\mathrm{CO_3}^{2-}\right]}$ .

Where

$a_{\mathrm{H_2CO_3}}$ , $a_{\mathrm{H^{+}}}$ , and $a_{\mathrm{CO_3}^{2-}}$ denote the activities of the three species, and
$[\mathrm{H_2CO_3}]$ , $\left[\mathrm{H^{+}}\right]$ , and $\left[\mathrm{CO_3}^{2-}\right]$ denote the concentrations of the three species.

Explanation:

<h3>Equilibrium Constant Expression</h3>

The equilibrium constant expression of a (reversible) reaction takes the form a fraction.

Multiply the activity of each product of this reaction to get the numerator. $\rm H_2CO_3\; (aq)$ is the only product of this reaction. Besides, its coefficient in the balanced reaction is one. Therefore, the numerator would simply be $\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)$ .

Similarly, multiply the activity of each reactant of this reaction to obtain the denominator. Note the coefficient " $2$ " on the product side of this reaction. $\rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq)$ is equivalent to $\rm H^{+}\, (aq) + H^{+}\, (aq) + {CO_3}^{2-}\, (aq)$ . The species $\rm H^{+}\, (aq)$ appeared twice among the reactants. Therefore, its activity should also appear twice in the denominator:

$\left(a_{\mathrm{H^{+}}}\right)\cdot \left(a_{\mathrm{H^{+}}}\right)\cdot \, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right = \left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right$ .

That's where the exponent " $2$ " in this equilibrium constant expression came from.

Combine these two parts to obtain the equilibrium constant expression:

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \quad\begin{matrix}\leftarrow \text{from products} \\[0.5em] \leftarrow \text{from reactants}\end{matrix}$ .

<h3 /><h3>Equilibrium Constant of Concentration</h3>

In dilute solutions, the equilibrium constant expression can be approximated with the concentrations of the aqueous " $(\rm aq)$ " species. Note that all the three species here are indeed aqueous. Hence, this equilibrium constant expression can be approximated as:

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{\left[\mathrm{H_2CO_3\, (aq)}\right]}{\left[\mathrm{H^{+}\, (aq)}\right]^2\cdot \left[\mathrm{{CO_3}^{2-}\, (aq)}\right]}$ .

8 0

3 years ago

Citric acid (H3 C6 H5 O7) is a product of the fermentation of sucrose (C12 H22 O11) in air.

stiks02 [169]

Answer:

960.6175

Explanation:

The balanced equation shows that for every mole of sucrose used, 2 moles of citric acid are produced.

So 2.5 moles of C12H22O11 will produce 2*2.5 moles of Citric acid.

So 2.5 moles of C23H22O11 will produce 5 moles of Citric Acid.

The next step is to figure out the molar mass of Citric Acid.

There are

Hydrogen: H3 + H5 moles of hydrogen in 1 mole of Citric Acid = 8 mol H

Carbon: 1 mol of Citric Acid has 6 moles of Carbon in it = 6 mol C

Oxygen: 1 mol of Citric Acid has 7 moles of Oxygen it it =7 mol O

Hydrogen has a mass of 1 so Citric Acid has 1 * 8 grams H 8

Carbon has a mass of 12 so Citric Acid has 12 * 6 grams C 72

Oxygen has a mass of 16 so Citric Acid has <u>16 * 7 grams O 112</u>

Total grams for 1 mol Citric Acid 192

The official published mass of Citric Acid is 192.1235 so because I used rounded numbers I'm a little off. It is important for any chemistry student student to know where the 192.1235 came from rather than exactly why I'm a little out. Every periodic table lists different masses for each element. If you want a better number, you will have to consult your own periodic table.

1 mol of Citric Acid = 192 grams

5 mols of Citric Acid = x

Cross multiply

1 * x = 5 * 192

x = 960 grams.

If you want to use the published amount then the answer will be 5 * 192.1235 = 960.6175

8 0

3 years ago