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EleoNora [17]
3 years ago
7

Hydrocarbons are compounds that are composed of carbon and hydrogen. Which statement does the law of definite proportions suppor

t?
A. Each sample of a single hydrocarbon has the same mass ratio of carbon to hydrogen
B. The mass ratio of carbon to hydrogen is the same for each hydrocarbon
C. No to hydrocarbons have the same mass ratio of carbon to hydrogen
Each hydrocarbon can have several mass ratio of carbon to hydrogen depending on its source
Chemistry
1 answer:
lora16 [44]3 years ago
6 0
The correct option is A.
The law of definite proportion states that a given chemical compound always contains its component elements in fixed ratio irrespective of the source they are obtain from.
For instance, water is made up of two atoms of hydrogen and one atom of oxygen. Water can not be formed in any other way. Combining the oxygen and the hydrogen in another ratio will give us another thing not water.
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In the process of eukaryotic pre-mRNA splicing, how is the lariat intermediate formed?
Aleks [24]

Answer:

C

Explanation:

The Eukaryotic pre-mRNA receives a 5' cap and 3'poly(A) tail before Introns are removed and the mRNA is considered ready for translation.

4 0
3 years ago
Can you help me with this with a solution
LuckyWell [14K]

The complete table is inserted.

A table is given,

Formulas used:

pH=  -log(H⁺)

pOH=  -log(OH⁻)

pH+ pOH=14

Calculations:

For A: (H⁺)=2×10⁻⁸M

Using the pH formula:

pH=  -log(H⁺)=-log(2×10⁻⁸)=7.69

pOH=14 - 7.69=6.3

Calculating OH concentration,

pOH=  -log(OH⁻)

6.3= -log(OH⁻)

(OH⁻)=5.011×10⁻⁷M

Hence, the nature of A is basic.

Similarily,

For B,

(OH⁻)=1×10⁻⁷

Using the pH formula:

pOH=  -log(OH⁻)= -log(1×10⁻⁷)=7

pH=14-7=7

Calculating H concentration,

pH=  -log(H⁺)

7= -log(H⁺)

(H⁺)=1×10⁻⁷M

Hence, the nature of B is neutral.

Similarily,

For C,

pH=12.3

Using the pH formula:

pOH=14-12.3=1.7

Calculating H concentration,

pH=  -log(H⁺)

12.3= -log(H⁺)

(H⁺)=5.011×10⁻¹³M

Calculating OH concentration,

pOH=  -log(OH⁻)

1.7= -log(OH⁻)

(OH⁻)=1.99×10⁻²M

Hence, the nature of C is Basic.

Similarily,

For D,

pOH=6.8

Using the pH formula:

pH=14-6.8=7.2

Calculating H concentration,

pH=  -log(H⁺)

7.2= -log(H⁺)

(H⁺)=6.309×10⁻⁸M

Calculating OH concentration,

pOH=  -log(OH⁻)

6.8= -log(OH⁻)

(OH⁻)=1.58×10⁻⁷M

Hence, the nature of D is basic.

Learn more about the acid and bases here:

brainly.com/question/16189013

#SPJ10

3 0
2 years ago
Provide a stepwise synthesis of 1-cyclopentylethanamine using the Gabriel synthesis. Collapse question part Testbank, Question 0
ra1l [238]

Answer:

Azide synthesis is the first method on the table of synthesis of primary amines. The Lewis structure of the azide ion, N3−, is as shown below.

an azide ion

An “imide” is a compound in which an N−−H group is attached to two carbonyl groups; that is,

imide linkage

You should note the commonly used trivial names of the following compounds.

phthalic acid, phthalic anhydride, and phthalimide

The phthalimide alkylation mentioned in the reading is also known as the Gabriel synthesis.

If necessary, review the reduction of nitriles (Section 20.7) and the reduction of amides (Section 21.7).

Before you read the section on reductive amination you may wish to remind yourself of the structure of an imine (see Section 19.8).

The Hofmann rearrangement is usually called the Hofmann degradation. In a true rearrangement reaction, no atoms are lost or gained; however, in this particular reaction one atom of carbon and one atom of oxygen are lost from the amide starting material, thus the term “rearrangement” is not really appropriate. There is a rearrangement step in the overall degradation process, however: this is the step in which the alkyl group of the acyl nitrene migrates from carbon to nitrogen to produce an isocyanate.

Explanation:

8 0
3 years ago
True or false: chemical reactions happen and compounds form because they're trying to get 6 electrons in their outer orbitals
wel

Answer:

False

Explanation:

The octet rule forms the basis for chemical reactions. The octet rule states that; an atom is only stable when it has eight electrons around its outermost shell.

This implies that the driving force behind chemical reaction is the attainment of an octet structure(eight electrons in the outermost shell of each of the bonding atoms).

An atom that has only six electrons in its outermost shell is not yet stable according to the demand of the octet rule. Hence, the statement "chemical reactions happen and compounds form because they're trying to get 6 electrons in their outer orbitals" is false.

7 0
3 years ago
Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe
baherus [9]

Answer:

The equilibrium constant is  K =0.02867

The equilibrium pressure for oxygen gas is  P_{O_2} =  0.09367\  atm

Explanation:

  From the question we are told that

       The equation of the chemical reaction is

                    M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}

   The Gibbs free energy forM_3 O_4_{(s)} is  \Delta G^o_{1}  = -8.80 \ kJ/mol

  The Gibbs free energy forM{(s)} is  \Delta G^o_{2}  = 0 \ kJ/mol

    The Gibbs free energy forO_2{(s)} is  \Delta G^o_{3}  = 0 \ kJ/mol

The Gibbs free energy of the reaction is mathematically represented as

         \Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r

         \Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)

Substituting values

From the balanced equation

         \Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]

        \Delta G^o_{re} = 8.80 kJ/mol =8800J/mol

The Gibbs free energy of the reaction can also be represented mathematically as

           \Delta G^o_{re} = -RTln K

Where R is the gas constant with a value of  R = 8.314 J/mol \cdot K

             T is the temperature with a given value  of  T = 298 K

             K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction  is mathematically represented as

                          K_p =[ P_{O_2}]^{\frac{3}{2} }

Where   [ P_{O_2}] is the equilibrium pressure of oxygen

         The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the  second equation of the Gibbs free energy of the reaction

 

           K = e^{- \frac{\Delta G^o_{re}}{RT} }

Substituting values

           K= e^{\frac{8800}{8.314 * 298} }

            K =0.02867

Now substituting this into the equation above to obtain the equilibrium of oxygen

           0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}

multiplying through by 1 ^{\frac{2}{3} }

        P_{O_2} =  [0.02867]^{\frac{2}{3} }

        P_{O_2} =  0.09367\  atm

       

3 0
3 years ago
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