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3 years ago

How do scientists test their ideas

Chemistry

1 answer:

-Dominant- [34]3 years ago

5 0

I think it’s B not rlly sure

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Elaborate on the difference in natural occurrences between fission and fusion reactions. A) Neither fission nor fusion reactions

vazorg [7]

Okay then I would go with choice B since fusion takes place in the sun which is a giant star.

8 0

3 years ago

Read 2 more answers

Molecular iodine, I2 (g), dissociates into iodine atoms at 545K with a first order rate constant of 0.344 1/s. If you start with

Dennis_Churaev [7]

Answer:

0.00915 M of $I_2$ remain after 5.16 seconds.

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = $0.344$ s⁻¹

Initial concentration $[A_0]$ = 0.054 M

Final concentration $[A_t]$ = ? M

Time = 5.16 s

Applying in the above equation, we get that:-

$[A_t]=0.054e^{-0.344\times 5.16}\ M=\frac{1\times \:0.054}{e^{1.77504}}\ M=0.00915\ M$

0.00915 M of $I_2$ remain after 5.16 seconds.

3 0

3 years ago

Which of the following would dissolve in water? CCl4 LiCl CH4 PCl6

babymother [125]

Like dissolves like
so water is polar

CCl4 is nonpolar
LiCl is polar
CH4 is nonpolar
PCl6 is nonpolar

so LiCl would dissolve

7 0

3 years ago

Read 2 more answers

In certain conditions, the equilibrium concentrations for components in the reaction 2NOCl(g) 2NO(g) + Cl2(g) are [NO] = 0.02 M,

Sonja [21]

Keq= (products)/ (reactants)

Keq= ( [NO]^2 x [Cl2]) / ( [NOCl]^2)

Keq= ( (0.02)^2 x (0.01) ) / (0.5)^2= 1.6 x 10-5

8 0

3 years ago

What is the percent composition of NaHCO3?

Katarina [22]

Answer:- Na = 27.37%, H = 1.20%, C = 14.30% and O = 57.14%

Solution:- For the percentage composition of a compound, the atomic mass of each atoms times its subscript is divided by the molar mass of the compound and multiplied by 100.

The given compound is $NaHCO_3$ .

mass of Na = 22.99 g

mass of H = 1.008 g

mass of C = 12.01 g

mass of O = 3(16) = 48 g

Molar mass of compound = 22.99 g + 1.008 g + 12.01 g + 48 g = 84.008 g

percentage of Na = $(\frac{22.99}{84.008})100$

= 27.37%

percentage of H = $(\frac{1.008}{84.008})100$

= 1.20%

percentage of C = $(\frac{12.01}{84.008})100$

= 14.30%

percentage of O = $(\frac{48}{84.008})100$

= 57.14%

5 0

3 years ago