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3 years ago

Balanced symbol equation for reaction between sodium carbonate solution and dilute sulphuric acid?

Chemistry

2 answers:

Dmitry [639]3 years ago

3 0

Answer:

The molecular equation for the reaction betweensodium carbonate and sulfuric acid is: 1. Na2CO3(aq)+H2SO4(aq)→Na2SO4(aq)+CO2(g)+H2O(l) N a 2 C O 3 ( a q ) + H 2 S O 4 ( a q ) → N a 2 S O 4 ( a q ) + C O 2 ( g ) + H 2 O ( l ) .

Explanation:

CaHeK987 [17]3 years ago

3 0

Explanation:

H2SO4 + Na2CO3 → Na2SO4 + H2O + CO2

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The diagram shows two molecules (A and B) with different

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Answer: chromatography

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3 years ago

What is the name of this acid: H3P ?

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Answer:

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3 years ago

Write the electron configuration for Silicon

morpeh [17]

Answer:

Si₁₄ = 1s² 2s² 2p⁶ 3s² 3p²

Explanation:

Silicon is present in group 14 of periodic table.

It is blue-gray color metalloid.

Its atomic mass is 28 g/mol.

It is mostly used to make allow.

The most important alloy are Al-Si and Fe-Si. These are used to make different machine tools, transformer plates, engine etc.

Its atomic number is 14 and electronic configuration can be written as,

Electronic configuration:

Si₁₄ = 1s² 2s² 2p⁶ 3s² 3p²

The noble gas electronic configuration or abbreviated electronic configuration can also be written.

Si₁₄ = [Ne] 3s² 3p²

The atomic number of neon is 10. Its electronic configuration is,

Ne₁₀ = 1s² 2s² 2p⁶

That's why we write [Ne] for 1s² 2s² 2p⁶ in abbreviated electronic configuration of Si.

5 0

3 years ago

What is the net ionic equation of the reaction of mgcl2 with naoh?

AveGali [126]

Answer:

The net ionic equation will be MgCl₂ + 2 NaOH → Mg(OH)₂ + 2 NaCl

Explanation:

Ionization of MgCl₂ is as follows

MgCl₂ → Mg²⁺ + 2 Cl⁻

Ionization of NaOH is as follows

NaOH → Na⁺ + OH⁻

It is a one type of substitution reaction where OH⁻ combined with Mg²⁺ to give magnesium hydroxide .

On the other hand Cl⁻ combined with Na⁺ to give sodium chloride as product.

Using proper stoichiometry to balanced the number of atoms in both side .

7 0

3 years ago

The two isotopes of chlorine are LaTeX: \begin{matrix}35\\17\end{matrix}Cl35 17 C l and LaTeX: \begin{matrix}37\\17\end{matrix}C

Kay [80]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 35 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 37 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.45 amu

Putting values in equation 1, we get:

$35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.775\times 100=77.5\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.775)=0.225\times 100=22.5\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

6 0

3 years ago