1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Radda [10]
3 years ago
15

A good hypothesis must be which of the following?

Chemistry
2 answers:
yanalaym [24]3 years ago
8 0
Testable, that is always required for a hypothesis. Otherwise, how can you prove it?
motikmotik3 years ago
4 0
Testable is the answer
hope it helps
You might be interested in
The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed
Oksanka [162]

Answer:

(a) The rate of disappearance of O_{2} is: 4.65*10^{-5} M/s

(b) The value of rate constant is: 0.83036 M^{-2}s^{-1}

(c) The units of rate constant is:  M^{-2}s^{-1}

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

2NO(g)+O_{2}->2NO_{2}

rate = -\frac{1}{2} \frac{d}{dt}[NO] = -\frac{d}{dt}[O_{2}] = \frac{1}{2}\frac{d}{dt}[NO_{2}] -----(1)

According to the question, the reaction is second order in NO and first order in  O_{2}.

Then we can say that, rate = k[NO]^{2}[O_{2}] -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

-\frac{d}{dt}[NO] = 9.3*10^{-5} M/s.

(a) From (1), we can get the rate of disappearance of O_{2}.

    Rate of disappearance of  O_{2} = -\frac{d}{dt}[O_{2}] = (0.5)*(9.3*10^{-5}) M/s = 4.65*10^{-5} M/s.

(b) The rate of the reaction can be obtained from (1).

    rate = -\frac{1}{2} \frac{d}{dt}[NO] = (0.5)*(9.3*10^{-5})

    rate = 4.65*10^{-5} M/s

   The value of rate constant can be obtained by using (2).

    rate constant = k = \frac{rate}{[NO]^{2}[O_{2}]}

    k = \frac{4.65*10^{-5}}{(0.040)^{2}(0.035)} = 0.83036 M^{-2}s^{-1}

(c) The units of the rate constant can be obtained from (2).

    k = \frac{rate}{[NO]^{2}[O_{2}]}

    Substituting the units of rate as M/s and concentrations as M, we get:

\frac{Ms^{-1} }{M^{3}} = M^{-2}s^{-1}

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

     rate\alpha [NO]^{2}

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of (1.8)^{2} = 3.24

     

5 0
3 years ago
The following reaction produces ethanoic acid (CHACOOH) from methanol (CH3OH) and carbon
tigry1 [53]

Answer:

\boxed{\text{300 g}}

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r:         32                          60

           CH₃OH + CO ⟶ CH₃COOH

m/g:        160

(a) Moles of CH₃OH

\text{Moles of CH$_{3}$OH} = \text{160 g CH$_{3}$OH }\times \dfrac{\text{1 mol CH$_{3}$OH }}{\text{32 g CH$_{3}$OH}}= \text{5.00 mol CH$_{3}$OH}

(b) Moles of CH₃COOH

\text{Moles of CH$_{3}$COOH} = \text{5.00 mol CH$_{3}$OH } \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{1 mol CH$_{3}$OH }} = \text{5.00 mol CH$_{3}$COOH}

(c) Mass of CH₃COOH

\text{Mass of CH$_{3}$COOH} =\text{5.00 mol CH$_{3}$COOH} \times \dfrac{\text{60 g CH$_{3}$COOH}}{\text{1 mol CH$_{3}$COOH}} = \textbf{300 g CH$_{3}$COOH}\\\\\text{The maximum mass of ethanoic acid that can be produced is $\boxed{\textbf{300 g}}$}

3 0
4 years ago
What effect does a quadrupling of the mass have upon the acceleration of the object
netineya [11]
The state in which all of the external forces acting upon an object are balanced; there is no acceleration. friction ..... quadrupling. doubling distance and quadrupling mass has the overall effect of the force
8 0
3 years ago
If an element gives away an electron, will it form a positive ion or a negative<br> ion?
Elan Coil [88]

Answer:

posotive

Explanation:

electrons give off negative pulse so take one away

8 0
4 years ago
Calculate the molar concentration of the Cl⁻ ions in 0.92 M MnCl2(aq).
scZoUnD [109]
MnCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of MnCl2 that dissolves.
MnCl2(s) --> Mn+(aq) + 2 Cl⁻(aq)
            [Cl⁻] = 0.92 mol MnCl2/1L × 2 mol Cl⁻ / 1 mol MnCl2 = 1.8 M
The answer to this question is [Cl⁻] = 1.8 M
4 0
3 years ago
Other questions:
  • Using the following combustion reaction, describe how it follows the laws of conservation of mass and energy. CH 4 +2O 2 CO 2 +2
    15·1 answer
  • What physical properties affect an object’s gravitational pull?
    9·1 answer
  • A glass of water contains 1.2 x 1024 molecules of water. An additional 4.0 moles of water are poured into the glass. How many mo
    15·1 answer
  • This type of triglyceride contains more than one double bond in the fatty acid carbon atoms
    15·1 answer
  • Sedimentary rocks formed from the remains of once-living things are
    10·2 answers
  • HELP Consider this equation...
    11·2 answers
  • Help on science questions please!!!
    13·1 answer
  • What will happen to earth if earth stop spinning?
    15·1 answer
  • Why are polar solutions not electrolytic?
    14·1 answer
  • Chromium has four naturally occurring isotopes. cr-50 (mass of 49.946 amu and 4.35% abundance); cr-52 (mass of 51.941 amu and 83
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!