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3 years ago

If there is only one type of cation that can be formed from an element, what do we do with its name?

Chemistry

1 answer:

KatRina [158]3 years ago

6 0

Answer:

Magnesium, Mgstart text, M, g, end text, is a group 2 element that will form 2+ cations. Because it usually forms cations of only one type, we don't need to specify its charge. We can simply refer to the cation in the ionic compound as magnesium. ... Therefore, the name for the compound is magnesium phosphide.

Explanation:

(this may or not be correct)

-also i finished that drawing you wanted :p-

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Write a balanced chemical equation, including physical state symbols, for the decomposition of solid calcium carbonate (CaCO_3)

nadya68 [22]

Answer:

CaCO₃(s) → CaO(s) + CO₂(g)

Explanation:

The decomposition reaction always make two compounds from one.

The products always have simpler chemical structure, originated from a determined compound. This can happens spontaneously or by a third party.

A notable example of decomposition is hydrolysis. As for example the case of water, which decomposes and generates oxygen and hydrogen gas

2H₂O (l) → 2 H₂ (g) + O₂ (g)

In this case, the calium carbonate decomposes into CaO and CO₂

These two, are the products of the decomposition.

Of course, the unique reactant is the Calcium Carbonate

The balanced equation is:

CaCO₃(s) → CaO(s) + CO₂(g)

6 0

3 years ago

Read 2 more answers

A solution with a ph of 4 has _________ the concentration of h+ present compared to a solution with a ph of 5.

Marina CMI [18]

A solution with a pH of 4 has ten times the concentration of H⁺ present compared to a solution with a pH of 5.
pH is a numeric scale for the acidity or basicity of an aqueous solution. It is the negative of the base 10 logarithm of the molar concentration of hydrogen ions.
[H⁺] = 10∧-pH.
pH = 4 → [H⁺]₁ = 10⁻⁴ M = 0,0001 M.
pH = 5 → [H⁺]₂ = 10⁻⁵ M = 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 0,0001 M / 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 10.

7 0

3 years ago

A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so

Xelga [282]

Answer : The equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

Explanation :

First we have to calculate the initial moles of $Fe^{3+}$ and $SCN^-$ .

$\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}$

$\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol$

and,

$\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}$

$\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol$

The given balanced chemical reaction is,

$Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$

Since 1 mole of $Fe^{3+}$ reacts with 1 mole of $SCN^-$ to give 1 mole of $FeSCN^{2+}$

The limiting reagent is, $SCN^-$

So, the number of moles of $FeSCN^{2+}$ = 0.0020 mmole

Now we have to calculate the concentration of $FeSCN^{2+}$ .

$\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M$

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution

l = path length

$\epsilon$ = molar absorptivity coefficient

$\epsilon$ and l are same for stock solution and dilute solution. So,

$\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}$

For trial solution:

The equilibrium concentration of $SCN^-$ is,

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{initial}$ = 0.00050 M

Now calculate the $[FeSCN^{2+}]$ .

$C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M$

Now calculate the concentration of $SCN^-$ .

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)$

$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

Therefore, the equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

5 0

4 years ago

When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV) the energy of the phot

viktelen [127]

Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

First we have to calculate the $'n^{th}'$ orbit of hydrogen atom.

Formula used :

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number of hydrogen atom = 1

Energy of n = 1 in an hydrogen atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6eV$

Energy of n = 2 in an hydrogen atom:

$E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV$

Energy change transition from n = 1 to n = 3 occurs.

Let energy change be E.

$E=E_-E_3=(-13.6eV)-(-1.51eV)=-12.1eV$

The negative sign indicates that energy of the photon emitted.

Thus, the energy of the photon emitted is, -12.1 eV

3 0

3 years ago

How can constraints be used to help define the problem?

vitfil [10]

Answer:

Constraints are restrictions that need to be placed upon variables

Explanation:

Constraints are restrictions (limitations, boundaries) that need to be placed upon variables used in equations that model real-world situations. It is possible that certain solutions which make an equation true mathematically, may not make any sense in the context of a real-world word problem.

6 0

3 years ago