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yanalaym [24]
2 years ago
3

To prevent a condition called the “bends”, deep sea divers breathe a mixture containing, in mole percent, 10.0% O2, 10.0% N2 , a

nd 80.0% He.
a. Calculate the molar mass of this mixture.
b. What is the ratio of the density of this gas to that of pure Oxygen?
Chemistry
1 answer:
trapecia [35]2 years ago
7 0
<span><span>1.       </span><span>In order to prevent the condition called “bends”. He needs to calculate the Molar Mass of the mixture of 10.0% 02, 10.0 % N2 and 80 % He.
So the Answer is letter A. Calculate the molar mass of this mixture.
How to calculate:
Given Numbers : 10.0 % O2
                              10.0% N2
                               80 % He
Molar Mass = (.10 x 32g / mol.) + (.10 x 28g / mol.) + (.80 x 4g / mol.)
                          => 9.2g / mol.</span></span>



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Pure platinum is too soft to be used in jewelry because it scratches easily. To increase the hardness so that it can be used in
Zina [86]

Answer:

percentage mass of platinum in the alloy ≈ 90.60 %

Explanation:

The alloy is 8.528 g sample of an alloy containing platinum and cobalt . The alloy react with excess nitric acid to form cobalt(ii) nitrate . Platinum is resistant to acid so it will definitely not react with the acid only the cobalt metal in the alloy will react with the acid.

The chemical reaction can be represented as follows:

Co (s) + HNO₃ (aq) → Co(NO₃)₂ (aq) +  NO₂ (l) + H₂O (l)

The balanced equation

Co (s) + 4HNO₃ (aq) → Co(NO₃)₂(aq) + 2NO₂ (l) + 2H₂O (l)

Cobalt is the limiting reactant

atomic mass of cobalt = 58.933 g/mol

Molar mass of Co(NO₃)₂ = 58.933 + 14 × 2 + 16 × 6 = 58.933 + 28 + 96 = 182.933  g

58.933 g of cobalt produce 182.933  g of  Co(NO₃)₂

? gram of cobalt will produce 2.49 g of Co(NO₃)₂

cross multiply

grams of cobalt that will react = (58.933 × 2.49)/182.933

grams of cobalt that will react = 146.74317000/182.933

grams of cobalt that will react= 0.8021689362 g

grams of cobalt that will react = 0.802 g

mass of platinum in the alloy  = 8.528 g - 0.802 g = 7.726 g

percentage  mass of platinum in the alloy = 7.726/8.528 × 100 = 772.600/8.528 = 90 .595 %

percentage mass of platinum in the alloy ≈ 90.60 %

6 0
2 years ago
The compound AX2 decomposes according to the equation, 2 AX2(g) =&gt; 2 AX(g) + X2(g). In one experiment, AX2 was measured at va
Karolina [17]

Answer:

0.0011 mol/L.s

Explanation:

The average rate of disappearing of the reagent is the variation of the concentration of it divided by the time that this variation is being measured. The reaction rate, is proportional to the coefficient of the substance, so, for a generic reaction:

aA + bB --> cC + dD

rate = -(1/a)Δ[A]/Δt = -(1/b)Δ[B]/Δt = (1/c)Δ[C]/Δdt = (1/d)Δ[D]/dt

The minus sign is because of the reagent is desapering, so:

rate = -(1/2)*(0.0209 - 0.0300)/(10 - 6)

rate = 0.0011 mol/L.s

8 0
2 years ago
The combustion of 0.374 kg of methane in the presence of excess oxygen produces 0.983 kg of carbon dioxide. What is the percent
anyanavicka [17]
Assuming that the combustion formula is
CH4 + 2O2 --> 2H2O + CO2<span>,

That means for every 1 molecule of methane(CH4) there will be one molecule of carbon dioxide(</span>CO2) produced. Methane molecular weight 16, carbon dioxide molecular weight is 44. Then the percent yield should be:
1 * (0.374/ 16) /(0.983/44)= 0.374*44/ 0.983 * 16= 104.6%

You sure the number is correct? Percent yield should not exceed 100%
5 0
2 years ago
A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.314 kg of water?
Mariulka [41]

Answer:

g NaCl = 424.623 g

Explanation:

<em>C</em> NaCl = 3.140 m = 3.140 mol NaCl / Kg solvent

∴ solvent: H2O

∴ mass H2O = 2.314 Kg

mol NaCl:

⇒ mol NaCl = (3.140 mol NaCl/Kg H2O)×(2.314 Kg H2O) = 7.266 mol NaCl

∴ mm NaCl = 58.44 g/mol

⇒ g NaCl = (7.266 mol NaCl)×(58.44 g/mol) = 424.623 g NaCl

5 0
2 years ago
Using the reaction below: 2 CO2(g) + 2 H2O(l) → C2H4(g) + 3 O2(g) ΔHrxn= +1411.1 kJ What would be the heat of reaction for this
maw [93]

Answer:  d) -705.55 kJ

Explanation:

Heat of reaction is the change of enthalpy during a chemical reaction with all substances in their standard states.

2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g) \Delta H=+1411.1kJ

Reversing the reaction, changes the sign of \Delta H

C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)

\Delta H=-1411.1kJ

On multiplying the reaction by \frac{1}{2} , enthalpy gets half:

0.5C_2H_4(g)+1.5O_2(g)\rightarrow CO_2(g)+H_2O(l)\Delta H=\frac{1}{2}\times -1411.1kJ=-705.55kJ/mol

Thus the enthalpy change for the given reaction is -705.55kJ

7 0
2 years ago
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