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jeyben [28]
3 years ago
13

At 500.K and 1.00 atm pressure, 1.5 liters of pentane gas, C5H12, is mixed with 15 liters of oxygen gas. A complete combustion r

esults. How many liters of WATER vapor, measured at the same temperature and pressure, would be produced after the reaction?
Chemistry
1 answer:
____ [38]3 years ago
5 0

Answer:

1.5 litre of C5H12 produce 9 litre of water vapour;

Explanation:

First balance the combustion reaction;

C_5H_12 +8O_2 → 5CO_2 +6H_2O

from the balance reacion it is clearly that,

one mole of C5H12 reacts completely with 8 mole of O2 gas;

Hence

one litre of C5H12 reacts completely with 8 litre of O2 gas;

there fore 1.5 litre C5H12 needs 12 litre oxygen gas but we have 15 litre.

so,

C5H12 is a limiing reactant and O2 is a excess reactant.

so quantity of H2O depends on limiting reactant;

one litre of C5H12 produce 6 litre of water vapour;

therefore,

1.5 litre of C5H12 produce 9 litre of water vapour;

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Five students performed a Kjeldahl nitrogen analysis of a protein sample. The following weight % nitrogen values were determined
Sunny_sXe [5.5K]

Answer:

G_calculated = 1.756

The outlier should be rejected, as G_cal > G_tab (= 1.463) at 95 % confidence.

Explanation:

The Grubb's test is used for identifying an outlier in data, which is from the same population. For this, a statistical term, G, is calculated for the suspected outlier. If the calculated value is greater than the tabulated G value then the suspected value is rejected. This term is given as,

G_calculated = | suspect value - mean| / s

Here,  suspect value is 13.8, mean is to be taken of all the data (including suspected value). s is the standard deviation of the sample data.

s is calculated from the following formula:

s = (Σ(xi - x)²/(N-1))^1/2

Here, x is the mean, which is 15.24, xi is individual value and N is the total number of data (5).

From the above formula, s is found to be

Standard Deviation, s = 0.820

Now for G value,

G_calculated = | 13.8 - 15.24| / (0.820)

G_ calculated = 1.756

The tabulated G value at 95 % confidence and N -1 (5 - 1 = 4) degree of freedom is, 1.463.

As calculated G (1.756) is greater than the tabulated G (1.463), the value 13.8 is considered an outlier at 95 % confidence.  

3 0
3 years ago
Osmosis is the passage of water molecules through a semi-permeable membrane from a solution of ________ concentration to _______
grandymaker [24]

Osmosis is the passage of water molecules through a semi-permeable membrane from a solution of ___lower_____ concentration to ___higher_____ concentration.

3 0
3 years ago
Solid silver chloride, AgCI(s), decomposes to its elements when exposed to sunlight Write a balanced equation for this reaction
Mrac [35]

Answer: The coefficient in front of AgCl when the equation is properly balanced is 2.

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

Decomposition of silver chloride is represented as:

2AgCl(s)\rightarrow 2Ag(s)+Cl_2(g)

Thus the coefficient in front of AgCl when the equation is properly balanced is 2.

8 0
4 years ago
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vampirchik [111]

Answer:

its b alkyne

Explanation:

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8 0
3 years ago
Read 2 more answers
2NO + 3MnO2 + 4H â 2NO3- + 3Mn2 + 2H2O For the above redox reaction, assign oxidation numbers and use them to identify the eleme
mixer [17]

Answer:

Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).

Explanation:

Hello there!

In this case, according to the given redox reaction, we rewrite it as a convenient first step:

2NO + 3MnO_2 + 4H^+ \rightarrow 2NO_3^- + 3Mn^{2+} + 2H_2O

Next, we assign the oxidation numbers as follows:

2N^{2+}O^{2-} + 3Mn^{4+}O^{-2}_2 + 4H^+ \rightarrow 2(N^{5+}O^{2-}_3)^- + 3Mn^{2+} + 2H^+_2O^{2-}

Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).

Regards!

4 0
3 years ago
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