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3 years ago

What the resalut H2SO4 +CO3-=?

Chemistry

1 answer:

BartSMP [9]3 years ago

6 0

Answer:

H2SO4 + CO3 = H2CO3 + SO4

Explanation:

H2SO4 + CO3

This is the reaction of sulphuric acid and carbon trioxide.

Balanced reaction is;

H2SO4 + CO3 = H2CO3 + SO4

So they react to produce carbonic acid and sulfate.

You might be interested in

What is meant by reactive elements

Vlad1618 [11]

Answer; relative elements are most reactive elements and compounds may ignite spontaneously or explosively. They generally burn in water as well as the oxygen in the air

Explanation:

6 0

3 years ago

Which pair of atoms forms a nonpolar covalent bond? which pair of atoms forms a nonpolar covalent bond? na and cl c and o n and

adelina 88 [10]

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

Less than 0.4 then it is Non Polar Covalent

Between 0.4 and 1.7 then it is Polar Covalent

Greater than 1.7 then it is Ionic

For Na and Cl,

E.N of Chlorine = 3.16

E.N of Sodium = 0.93

________

E.N Difference 2.23 (Ionic Bond)

For C and O,

E.N of Oxygen = 3.44

E.N of Carbon = 2.55

________

E.N Difference 0.89 (Polar Covalent Bond)

For N and Cl,

E.N of Chlorine = 3.16

E.N of Notrogen = 3.04

________

E.N Difference 0.12 (Non-Polar Covalent Bond)

For B and O,

E.N of Oxygen = 3.44

E.N of Boron = 2.04

________

E.N Difference 1.40 (Polar Covalent Bond)

8 0

4 years ago

Read 2 more answers

At what pressure would 11.1 moles of a gas occupy 44.8 L at 300 K?

faltersainse [42]

Answer:

$P=6.10atm$

Explanation:

Hello,

In this case, we can study the ideal gas equation that relates temperature, volume, pressure and moles as shown below:

$PV=nRT$

Thus, since we are asked to compute the pressure y simply solve for it as follows:

$P=\frac{nRT}{V}=\frac{11.1mol*0.082\frac{atm*L}{mol*K}*300K}{44.8L}\\ \\P=6.10atm$

Best regards.

5 0

3 years ago

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

3 years ago

How many molecules are in 3.6 grams of NaCl? Question options:

Levart [38]

Answer:

$\boxed {\boxed {\sf 3.7 * 10^{22} \ molecules \ NaCl}}$

Explanation:

We are asked to find how many molecules are in 3.6 grams of sodium chloride.

<h3>1. Convert Grams to Moles </h3>

First, we convert grams to moles using the molar mass. These values are equivalent to atomic masses on the Periodic Table, but the units are grams per moles instead of atomic mass units. Look up the molar masses of the individual elements: sodium and chlorine.

Na: 22.9897693 g/mol
Cl: 35.45 g/mol

There are no subscripts in the chemical formula (NaCl), so we simply add the 2 molar masses.

NaCl: 22.9897693 + 35.45 = 58.4397693 g/mol

Now we will convert using dimensional analysis. First, set up a ratio using the molar mass.

$\frac {58.4397693 \ g \ NaCl}{ 1 \ mol \ NaCl}$

We are converting 3.6 grams to moles, so we must multiply the ratio by this value.

$3.6 \ g \ NaCl *\frac {58.4397693 \ g \ NaCl}{ 1 \ mol \ NaCl}$

Flip the ratio so the units of grams of sodium chloride cancel.

$3.6 \ g \ NaCl *\frac { 1 \ mol \ NaCl}{58.4397693 \ g \ NaCl}$

$3.6 *\frac { 1 \ mol \ NaCl}{58.4397693}$

$\frac { 3.6}{58.4397693} \ mol \ NaCl$

$0.06160188589 \ mol \ NaCl$

<h3>2. Convert Moles to Molecules </h3>

Next, we convert moles to molecules using Avogadro's Number. This is 6.022 × 10²³ and it tells us the number of particles (atoms, molecules, formula units, etc). In this case, the particles are molecules of sodium chloride. Let's set up another ratio.

$\frac {6.022 \times 10^{23} \ molecules \ NaCl}{ 1 \ mol \ NaCl}$

Multiply by the number of moles we calculated.

$0.06160188589 \ mol \ NaCl * \frac{6.022 \times 10^{23} \ molecules \ NaCl}{1 \ mol \ NaCl}$

The units of moles of sodium chloride cancel.

$0.06160188589 * \frac{6.022 \times 10^{23} \ molecules \ NaCl}{1 }$

$3.70966557*10^{22} \ molecules \ NaCl$

<h3>3. Round </h3>

The original measurement of grams (3.6) has 2 significant figures, so our answer must have the same. For the number we found, that is the tenths place. The 0 in the hundredth place tells us to leave the 7 in the tenth place.

$3.7 * 10^{22} \ molecules \ NaCl$

There are $3.7 * 10^{22} \ molecules \ NaCl$ in 3.6 grams and the correct answer is choice D.

5 0

3 years ago