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jasenka [17]
3 years ago
12

Convert 0.85 x 10-12 km to nm.

Chemistry
2 answers:
vitfil [10]3 years ago
8 0

Answer:

8,5 E10 nm

Explanation:

⇒ 0.85 E-2 Km * ( 1000 m / Km ) * ( 1 E9 nm / m ) = 8.5 E10 nm

⇒ 0.85 E-2 Km = 8.5 E10 nm

brilliants [131]3 years ago
7 0

Answer : The 0.85\times 10^{-12}km is equal to 0.85 nm.

Explanation :

The conversion used from kilometer (km) to nanometer (nm) is:

1km=10^{12}nm

As we are given that the distance 0.85\times 10^{-12}km. Now we have to calculate the distance in 'nm'.

As, 1km=10^{12}nm

So, 0.85\times 10^{-12}km=\frac{0.85\times 10^{-12}km}{1km}\times 10^{12}nm=0.85nm

Therefore, the 0.85\times 10^{-12}km is equal to 0.85 nm.

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Consider the following reaction at 298 K:
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Answer :

\Delta S_{sys} = -1622.8 J/K

\Delta S_{surr} = -94.6 J/K

\Delta S_{univ} = 0 J/K

Explanation :  Given,

\Delta H = -483.6 kJ

The given chemical reaction is:

2H_2(g)+O_2(g)\rightarrow 2H_2O(g)

First we have to calculate the value of \Delta S_{sys}.

\Delta S_{sys}=\frac{\Delta H}{T}

\Delta S_{sys}=\frac{-483.6kJ}{298K}

\Delta S_{sys}=-1.6228kJ/K=-1622.8J/K

Entropy of system = -1622.8 J/K

As we know that:

Entropy of system = -Entropy of surrounding = 1622.8 J/K

and,

Entropy of universe = Entropy of system + Entropy of surrounding

Entropy of universe = -1622.8 J/K + (1622.8 J/K)

Entropy of universe = 0

8 0
3 years ago
1.Build or draw the Lewis structure for each of the molecules listed below.
bagirrra123 [75]
This may help you

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explanation:

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6 0
2 years ago
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What is the concentration, in m/v percent, of a solution prepared from 50 g NaCl and 2.5 L of solution?
Serjik [45]
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4 0
3 years ago
Given that blood exerts the same osmotic pressure as a 0.15 M NaCl solution, which solution could be the hypertonic solution?/se
inna [77]

Answer:

a. 0.68 M NaCl solution

Explanation:

The tonicity of a solution can either be hypotonic, hypertonic and isotonic. Hypertonic solution is a solution which possesses a higher concentration of solute in relation to another solution.

According to this question, a solution is said to contain 0.15 M NaCl solute. This means that a solution that will be hypertonic to this solution will have a much more concentration of solute, which based on the options provided is the 0.68 M NaCl solution.

5 0
3 years ago
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