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3 years ago

5

What question could you ask about the components of two atoms to find out if the atoms are of the same elements or of different

elements

1 answer:

Temka [501]3 years ago

5 0

The number of protons and the electron configuration of each

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Aqueous sulfuric acid H2SO4 will react with solid sodium hydroxide NaOH to produce aqueous sodium sulfate Na2SO4 and liquid wate

allochka39001 [22]

Answer:

a) 2NaOH+H2SO4→Na2SO4+2H2O2NaOH+H2SO4→Na2SO4+2H2O

b) Số phân tử NaOH : Số phân tử H2SO4 = 2:1

Số phân tử NaOH : Số phân tử Na2SO4 = 2:1

Số phân tử NaOH : Số phân tử H2O = 2:2

Explanation:

8 0

2 years ago

Write ionic equations for any three of the following:

worty [1.4K]

1.

The ionic equation is:

H⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + NO₃⁻(aq) + H₂O(l)

HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H2O(l)

The required dissociations are

HNO₃(aq) → H⁺(aq) + NO₃⁻(aq)

NaOH(aq) → Na⁺(aq) + OH⁻(aq)

NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)

So, the ionic equation is

H⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + NO₃⁻(aq) + H₂O(l)

2.

The ionic equation is

H⁺(aq) + Cl⁻(aq) + K⁺(aq) + OH⁻(aq) → K⁺(aq) + Cl⁻(aq) + H₂O(l)

HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)

The required dissociations are

HCl(aq) → H⁺(aq) + Cl⁻(aq)

KOH(aq) → K⁺(aq) + OH⁻(aq)

KCl(aq) → K⁺(aq) + Cl⁻(aq)

So, the ionic equation is

H⁺(aq) + Cl⁻(aq) + K⁺(aq) + OH⁻(aq) → K⁺(aq) + Cl⁻(aq) + H₂O(l)

3.

The ionic equation is

2H⁺(aq) + SO₄²⁻(aq) + Mg²⁺(aq) + 2OH⁻(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

+ 2H₂O(l)

H₂SO₄(aq) + Mg(OH)₂(aq) → MgSO₄(aq) + 2H2O(l)

The required dissociations are

H₂SO₄(aq) → 2H⁺(aq) + SO₄²⁻(aq)

Mg(OH)₂(aq) → Mg²⁺(aq) + 2OH⁻(aq)

MgSO₄(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

So, the ionic equation is

2H⁺(aq) + SO₄²⁻(aq) + Mg²⁺(aq) + 2OH⁻(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

+ 2H₂O(l)

Learn more about ionic equations here:

brainly.com/question/11628165

5 0

2 years ago

Explain why the amino group of p-aminobenzoic acid does not participate in the reaction

lubasha [3.4K]

This is because amino group of p-aminobenzoic acid is an aniline and is less electrophilic than an alkyl amine.

<h3>What is an Aniline?</h3>

This is an aromatic amine which consists of a phenyl group attached to an amino group.

The amino group of p-aminobenzoic acid being an aniline makes it less electrophilic which is why an alkyl amine participates in the reaction instead.

Read more about Aniline here brainly.com/question/9982058

5 0

2 years ago

Pharmacists sometimes measure medicines in the unit of "grains," where 1 gr = 65 mg. The label on a bottle of aspirin reads "Asp

viva [34]

Answer:

325mg of Aspririn

Explanation:

First you should note the information that the problem gives you:

- The bottle of Aspirin has 5gr (grains)

- 1gr(grain) = 65mg (miligrams)

Also, the problem is asking about how many aspirin are in 5 gr (grains), so you should use a conversion factor, as follows:

-First you should put the quantity you need to convert:

$5grAspirin$

-Then you write the denominator of the conversion factor that must have the same units that you want to convert, in this case gr:

$5grAspirin*\frac{}{1grAspirin}$

-Then you write the numerator with the units that you want to obtain and the numerical equivalence between the units, in this case:

$5grAspirin*\frac{65mgAspririn}{1grAspirin}$

-Finally you multiply numerators and divide by denominators:

$5grAspirin*\frac{65mgAspririn}{1grAspirin}=325mgAspririn$

8 0

3 years ago

In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe

const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = $10^{-9} m$ .

Hence, 0.283 nm = $0.283 \times 10^{-9} m$

Formula for coulombic energy is as follows.

$U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}$

where, e = $1.6 \times 10^{-19}$ C

$\epsilon_{o}$ = $8.85 \times 10^{-12}$

$U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}$

= $1.423 \times 10^{-18} J$

As 1 eV = $1.6 \times 10^{-19} J$

So, 1 J = $\frac{1 eV}{1.6 \times 10^{-19}}$

Hence, U = $\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}$

= 8.9 eV

Also, 1 J = $\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}$

= $1.67 \times 10^{-27}$ kJ/mol

Therefore, U = $1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}$ kJ/mol

= $2.37 \times 10^{-45} kJ/mol$

7 0

3 years ago