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mixer [17]
3 years ago
12

You are given an unknown gaseous binary compound (that is, a compound consisting of two different elements). when 10.0 g of the

compound is burned in excess oxygen, 16.3 g of water is produced. the compound has a density 1.38 times that of oxygen gas at the same conditions of temperature and pressure. give a possible identity for the unknown compound.weibo.com/u/1157067277/home?topnav=1&wvr=6
Chemistry
1 answer:
Luda [366]3 years ago
8 0

Answer is: a possible identity for the unknown compound is C₃H₈.

m(binary compound) = 10.0 g.

m(H₂O) = 16.3 g; mass of water.

M(O₂) = 32 g/mol; molar mass of oxygen.

M(binary compound) = 1.38 · 32 g/mol.

M(binary compound) = 44.16 g/mol.

n(binary compound) = 10 g ÷ 44.16 g/mol.

n(binary compound) = 0.225 mol; amount of substance.

n(H₂O) = 16.3 g ÷ 18 g/mol.

n(H₂O) = 0.9 mol; amount of water.

m(H₂O) : n(binary compound) = 0.9 mol ÷ 0.225 mol.

m(H₂O) : n(binary compound) = 4 : 1.

Unknown cpmpound has 4 times more hydrogen than water, it has 8 hydrogen atoms.

Second element in compound is carbon:

M(X) = 44.16 g/mol - 8 · 1.01 g/mol.

M(X) = 36.08 g/mol ÷ 3.

M(C) = 12.01 g/mol.

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Arisa [49]

Answer:

Dilute

Explanation:

A concentrated solutions is a one which has relatively large amount of dissolved solute in the solution whereas a dilute solution is a one which has relatively lower concentration of dissolved solute.

In the given solution there is only 3.3% of solute. So, we can say that the given solution is a dilute solution. However, these terms are relative.

6 0
2 years ago
Please Help! Thank You
Ivanshal [37]

Explanation:

the answer is true because I had this question and got it right

3 0
2 years ago
I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol
galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = \Delta H_f_{(I_2)}=62.438 kJ/mol

Enthalpy of formation of chlorine gas = \Delta H_f_{(Cl_2)}=0 kJ/mol

Enthalpy of formation of ICl gas = \Delta H_f_{(ICl)}=17.78 kJ/mol

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]

=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol

Enthaply change when 1.62 moles of iodine gas recast:

\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ

Entropy of the surrounding = \Delta S^o_{surr}=\frac{\Delta H}{T}

=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0
3 years ago
2MG+O2 ...........2MGO
Galina-37 [17]
Mg + 1/2 O2 → MgO

1 mol = 24 g of Mg

X mol = 12 g of Mg

x = 0.5 moles of Mg

Mg :MgO = 1:1 (coefficient from equations using mole ratio)

So

0.5 moles of MgO

1 mol MgO = (24+16) g = 40 g

0.5 moles of MgO = 0.5 × 40

= 20 g of MgO produced
5 0
3 years ago
Which of the following examples is a molecule with a covalent bond?
serious [3.7K]

Answer:

B. CO

Explanation:

4 0
3 years ago
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