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2 years ago

In CaSO4, the oxidation number of Ca is that of S is and that of O is

Chemistry

1 answer:

Assoli18 [71]2 years ago

7 0

Answer: In CaF2, the oxidation number of Ca is +2

, and that of F is -1

. In H2SO4, the oxidation number of H is +1

, that of S is +6

, and that of O is -2

. In CaSO4, the oxidation number of Ca is +2

, that of S is +6

, and that of O is -2

. In HF, the oxidation number of H is +1

, and that of F is -1

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Natasha2012 [34]

Answer:

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2 years ago

How many moles of CaCO3 are needed to react with 12.5 mol SO2

grandymaker [24]

<h3><u>Answer;</u></h3>

= 12.5 Moles of CaSO3

<h3><u>Explanation</u>;</h3>

The reaction between CaCO3 and SO2 is given by the equation.

CaCO3(s) + SO2(g) → CaSO3(aq) + CO2(g)

The mole ratio between CaCO3 and SO2 is 1 : 1;

1 mole of CaCO3 reacts with 1 mole SO2 to form CaSO3 and CO2

Therefore;

<em>12.5 moles of SO2 will require 12.5 moles of CaSO3</em>

7 0

2 years ago

Which of these properties is the best one to use for identification of an element?

katovenus [111]

the number of protons in the atomic nucleus

3 0

2 years ago

The heat of vaporization of 1-pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K.mol. What is the approximate b

____ [38]

Answer:

Approximately 100 °C.

Explanation:

Hello,

In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:

$\Delta S_{vap}=\frac{\Delta H_{vap}}{T}$

We can solve for the temperature as follows:

$T=\frac{\Delta H_{vap}}{\Delta S_{vap}}$

Thus, with the proper units, we obtain:

$T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C$

Hence, answer is approximately 100 °C.

Best regards.

6 0

2 years ago

Element X has two isotopes. If 72.0% of the element has an isotope mass of 84.9 atomic mass units, and 28.0% of the element has

bija089 [108]

<h2>Answer:</h2>

Average atomic mass of an element is the sum of the masses of its isotopes each multiplied by its natural abundance

$\footnotesize \longrightarrow \: \rm Average \: atomic \: mass = \dfrac{ \sum\limits \: \% age \: of \: each \: isotope \times Atomic \: mass }{100} \\$

$\footnotesize \longrightarrow \: \rm Average \: atomic \: mass = \dfrac{ 72 \times84.9 + 28 \times 87 }{100} \\$

$\footnotesize \longrightarrow \: \rm Average \: atomic \: mass = \dfrac{ 6112.8 + 2436 }{100} \\$

$\footnotesize \longrightarrow \: \rm Average \: atomic \: mass = \dfrac{ 8548.8 }{100} \\$

$\footnotesize \longrightarrow \: \bf Average \: atomic \: mass = 85.488 \: amu \\$

8 0

2 years ago