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erica [24]
3 years ago
15

Write the symbol for the element which has the smallest atomic radius.c, al, li, ne.

Chemistry
1 answer:
enot [183]3 years ago
7 0
Ne because as we go in period atomic radius decreases
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HgS + O2 → HgO + SO2
Igoryamba

Answer:

2HgS + 3O2 → 2HgO + 2SO2

The coefficients are: 2, 3, 2, 2

Explanation:

HgS + O2 → HgO + SO2

The equation can be balance as follow:

Put 3 in front of O2 as shown below:

HgS + 3O2 → HgO + SO2

Now we can see that there are 6 atoms of O on the left side of the equation and a total of 3 atoms on the right side. It can be balance by putting 2 in front of HgO and SO2 as shown below:

HgS + 3O2 → 2HgO + 2SO2

Now we have 2 atoms of both Hg and S on the right side and 1atom each on the left. It can be balance by putting 2 in front of HgS as shown below:

2HgS + 3O2 → 2HgO + 2SO2

Now the equation is balanced.

The coefficients are: 2, 3, 2, 2

The law of conservation of mass(matter) states that matter(mass) can neither be created nor destroyed during a chemical reaction but changes from one form to another. An unbalanced equation suggests that matter has been created or destroyed. While a balanced equation proofs that matter can never be created but changes to different form. This is the more reason we have count the atoms of an element on both side of the equation to see if they are balanced irrespective of the new form they assume in the product

5 0
3 years ago
Determine the molality of an aqueous solution that is 16.5 percent urea by mass
yarga [219]

Take a hypothetical sample of exactly 100 grams of the solution.  

(16g urea) / (60.06 g urea/mol) = 0.2664 mol urea  

((100 g total) - (16g urea)) = 84.0 g H2O = 0.0840 kg H2O  

(0.2664  mol) /0.0840  (kg) = 3.17143mol/kg = 3.18m urea


7 0
3 years ago
The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C
avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

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

where,

q = heat gained = ?

c = specific heat = 5.20kJ/^oC

T_{final} = final temperature = 27.43^oC

T_{initial} = initial temperature = 22.93^oC

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

q=5.20kJ/^oC\times (27.43-22.93)^oC

q=23.4kJ

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

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

where,

\Delta H = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = \frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole

\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

\Delta H=\Delta U+\Delta n_gRT

or,

\Delta U=\Delta H-\Delta n_gRT

where,

\Delta H = change in enthalpy = -2805.8kJ/mol

\Delta U = change in internal energy = ?

\Delta n_g = change in moles = 0   (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = 27.43^oC=273+27.43=300.43K

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

\Delta U=\Delta H-\Delta n_gRT

\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K

\Delta U=-2805.8kJ/mol-0

\Delta U=-2805.8kJ/mol

Therefore, the internal energy change is -2805.8 kJ/mol

5 0
3 years ago
Nemistry I sem
GaryK [48]

Answer:

A

Explanation:

BF3 has a trigonal planar shape.

XeO3 has a trigonal pyramidal shape.

3 0
3 years ago
Protons are made up of _____ up quarks and one down quark
suter [353]

Answer:

two

Explanation:

7 0
3 years ago
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