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marshall27 [118]
3 years ago
5

An 18.5 g sample of tin (molar mass = 118.7) combines with

Chemistry
1 answer:
nasty-shy [4]3 years ago
5 0

Answer:

Option (B) SnS2

Explanation:

Data obtained from the question include:

Mass of Sn = 18.5 g

Mass of S = 10 g

The empirical formula of the compound can be obtained by doing the following:

Step 1:

Divide each element by their molar mass

Sn = 18.5/ 118.7 = 0.156

S = 10/ 32.07 = 0.312

Step 2:

Divide each by the smallest number

Sn = 0.156/0.156 = 1

S = 0.312/0.156 = 2

The empirical formula is SnS2

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When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t
OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

Mn(s)+2HCl(aq)\rightarrow  MnCl_2(aq)+H_2(g)

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

m=1.00 g/mL\times 100.0 mL = 100 g

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

q=m\times c\times (T_{final}-T_{initial})

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = 4.18 J/^oC

T_{final} = final temperature = 23.1^oC

T_{initial} = initial temperature = 28.9^oC

Now put all the given values in the above formula, we get:

q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC

q=2,242.4 J=2.242 kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = \frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol

\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

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The energy level can be defined as the fixed distances from the nucleus of an atom where electrons may be found. Each energy level is divided into some Subshells. These subshells are known as s-subshell, p-subshell, d-subshell, and f-subshell. This subshell contains some orbitals, these orbitals are the place where there is the maximum probability of getting the electrons. In one orbital, a maximum of two electrons can be present.

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