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Tasya [4]
3 years ago
5

A 42.0g sample of compound containing only C and H was analyzed. The results showed that the sample contained 36.0g of C and 6.0

g of H.
Which of the following questions about the compound can be answered using the results of the analysis?

A) What was the volume of the sample?
B) What is the molar mass of the compound?
C) What is the chemical stability of the compound?
D) What is the empirical formula of the compound?
Chemistry
1 answer:
user100 [1]3 years ago
5 0

Answer:

What is the empirical formula of the compound?

Explanation:

When the relative masses of elements in a hydrocarbon are given, it is possible to use this information to obtain the empirical formula by dividing the given masses of each element by the relative atomic masses of the element. The lowest ratio is now used to divide through to obtain the empirical formula of the compound.

The empirical formula only shows that ratio of atoms of each element present in the compound. From the information provided, the empirical formula of the compound is CH2. Hence the answer.

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How many moles of sodium hydroxide would react with 1 Mole of sulphuric acid?
Paul [167]

Answer:

Two moles.

Explanation:

Sulphuric (sulfuric) acid \rm H_2SO_4 is a diprotic acid. When one mole of \rm H_2SO_4 molecules dissolve in water, two moles of \rm H^{+} ions would be produced.

\rm H_2SO_4 \to 2\, H^{+} + {SO_4}^{2-}.

On the other hand, sodium hydroxide \rm NaOH is a monoprotic base. When one mole of \rm NaOH formula units dissolve in water, only one mole of hydroxide ions \rm OH^{-} would be produced.

\rm NaOH \to Na^{+} + OH^{-}.

Note that \rm H^{+} and \rm OH^{-} react at a one-to-one ratio:

\rm H^{+} + OH^{-} \to H_2O.

As a result, it would take 2\; \rm mol of \rm OH^{-} to react with the \rm 2\; mol of \rm H^{+} that was released when 1\; \rm mol of \rm H_2SO_4 is dissolved in water. Since one mole of \rm NaOH formula units could produce only one mole of \rm OH^{-}, it would take \rm 2\; mol of \rm NaOH formula units to produce that 2\; \rm mol of \rm OH^{-} for reacting with 1\; \rm mol of \rm H_2SO_4.

3 0
3 years ago
HELP! What is the percent composition of carbon in caffeine (C₈H₁₀N₄O₂)? Show your work
Dominik [7]

Caffeine has the following percent composition: carbon 49.48%, hydrogen 5.19%, oxygen 16.48% and nitrogen 28.85%. Its molecular weight is 194.19 g/mol.

8 0
3 years ago
What is the name for this type of reproduction?​
mario62 [17]

Answer:

----------------------------------------------

Reproduction is defined as the production of individuals of the same species, that is the next generation of the species.

----------------------------------------------

There are basically two types of reproduction - asexual and sexual.

----------------------------------------------

Asexual reproduction involves only one parent and the offspring is

genetically similar to the parent ...

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The different types of asexual reproduction are fission, budding, fragmentation, spore formation and vegetative propagation.

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Sexual reproduction is a type of reproduction that involves a complex life cycle in which a gamete (such as a sperm or egg cell) with a single set of chromosomes combines with another to produce a organism composed of cells with two sets of chromosomes .

Asexual reproduction is a type of reproduction which does not involve the fusion of gametes or change in the number of chromosomes. The offspring that arise by asexual reproduction from a single cell or from a multicellular organism inherit the genes of that parent.

Explanation:

Your very welcome!!! :) :) :)

4 0
3 years ago
The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th
MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

\frac{K_1}{K_2}=?

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

The expression used with catalyst and without catalyst is,

\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}

\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}

where,

K_2 = rate constant reaction -1

K_1 = rate constant reaction -2

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = 25^oC=273+25=298 K

Now put all the given values in this formula, we get

\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340

0.4284 is the ratio of the rate constants.

7 0
3 years ago
The reaction NA3PO4(aq) + 3 AgNO3(aq) → Ag3PO4() + 3NaNO 3( aq) is best classified as a(n)?
Natasha2012 [34]

Answer:

Double replacement reaction.

Explanation:

The Na and Ag atoms both (double) trade places (replacement) with each other.  

8 0
2 years ago
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