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stellarik [79]
3 years ago
5

if the observed test value of a hypothesis test is outside of the established critical value(s), a researcher would __________.

claim significant support for the hypothesis claim no significant support for the hypothesis restate the hypothesis more clearly repeat the experiment until support is found
Chemistry
2 answers:
romanna [79]3 years ago
8 0

Answer:

The correct answer is a

Explanation:

Makovka662 [10]3 years ago
6 0
The answer to this question is:

<span>If the observed test value of a hypothesis test is outside of the established critical value(s), a researcher would __________.
</span><span>"Claim significant support for the hypothesis"

Hoped This Helped, </span><span> Itsalishamariee
Your Welcome :)</span>
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A gas that exerts a pressure of
spayn [35]

Answer:

3.089 L

Explanation:

From the given information, provided that the no of moles and the temperature remains constant;

P_1 = 15.6 psi

V_1 = ???

P_2 = 25.43 psi

V_2 = 1.895 L

Using Boyle's law:

P_1V_1 =P_2V_2 \\ \\ V_1 = \dfrac{P_2V_2}{P_1} \\ \\  V_1 = \dfrac{25.43 \times 1.895}{15.6}  \\ \\ \mathbf{  V_1 = 3.089  \ L}

4 0
3 years ago
The reaction A(B) = 2B(g) has an equilibrium constant of K = 0.045. What is the equilibrium constant for the reaction B(g) =1/2A
Mamont248 [21]

Answer:

The  K_c for the reaction B(g) = \frac{1}{2}A will be 4.69.

Explanation:

The given equation is A(B) = 2B(g)

to evaluate equilibrium constant for B(g) = \frac{1}{2}A

            K_c=[B]^2[A]

                 = 0.045

The reverse will be 2B\leftrightharpoons A

Then,      K_c = \frac{[A]}{[B]^2}

                    =  \frac{1}{0.045}

                    = 22m^{-1}

The equilibrium constant for B(g) = \frac{1}{2}A will be

               K_c = \sqrt{K_c}

                    =\sqrt{22}

                    = 4.69

Therefore, K_c for the reaction B(g) = \frac{1}{2}A will be 4.69.

5 0
3 years ago
The rate constant for this second‑order reaction is 0.190 M − 1 ⋅ s − 1 0.190 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶product
Mamont248 [21]

Answer:

9.1 seconds

Explanation:

Given that for a second order reaction

1/[A]t = kt + 1/[A]o

Where [A]t= concentration at time = t= 0.340M

[A]o= initial concentration = 0.820M

k= rate constant for the reaction=0.190m-1s-1

t= time taken for the reaction (the unknown)

Hence;

(0.340)^-1 = 0.190×t + (0.820)^-1

t= (0.340)^-1 - (0.820)^-1/0.190

t= 9.1 seconds

Hence the time taken for the concentration to decrease from 0.840M to 0.340M is 9.1 seconds.

5 0
3 years ago
A newly discovered element, Y, has two naturally occurring isotopes. 87.8 percent of the sample is an isotope with a mass of 267
zloy xaker [14]
(0.878)(267.8) + (0.122)(269.9)=

268 u (three sig fig) 


Check my calculations
3 0
3 years ago
Read 2 more answers
Which molecules would most likely cause a liquid to have the lowest viscosity?
Alinara [238K]

The molecules with the lowest viscosity is the liquid with the least complex molecular structure. The structure affects the viscosity of the fluid because they are less compact and can flow freely.

8 0
3 years ago
Read 2 more answers
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