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inessss [21]
3 years ago
12

Juan wants to dilute a 5 molar solution of sulfuric acid down to a 1 molar solution. What would be the best way for him to do th

is?
A.
Add 400 mL of water to a beaker filled with 100 mL of sulfuric acid.
B. Add 400 mL of a base such as NaOH to 100 mL of sulfuric acid.
C. Add 100 mL of sulfuric acid to 400 mL of a base such as NaOH.
D.
Add 100 mL of sulfuric acid to a beaker filled with 400 mL of water.
Physics
2 answers:
Sophie [7]3 years ago
8 0
Add 100 ML of sulfuric acid to 400 ML of base as an 80H B.
eduard3 years ago
6 0
The answer is D. Add 100 mL of sulfuric acid to a beaker filled with 400mL of water
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Answer:

hi sandra!!

Explanation:

the number 3 is the correct!!

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4 0
3 years ago
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Will give brainliest
Nostrana [21]

Answer:

I think 18 not entirely sure though

Explanation:

Group 18 are the noble group so 16 + 2 = 18

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3 years ago
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On an asteroid, the density of dust particles at a height of 3 cm is ~30% of its value just above the surface of the asteroid. A
Anvisha [2.4K]

From the law of atmosphere

N_v(y) = n_0*e^{-\frac{mgy}{Kb*T}}

Where

n_0 = constant and is number density where the height y = 0cm

n_V = Number density at height y=3cm

Kb = Boltzmann constant = 1.38*10^{-23}J/K

T=20K

m = 10^{-19}kg

Re-arranging the equation to have the value of the gravity,

\frac{N_v(y)}{n_0} = e^{-\frac{mghy}{KbT}}

ln(\frac{N_v(y)}{n_0}) = -\frac{mgy}{KbT}

Since it is 30% of value above surface, therefore N_v = 0.3n_0

ln(\frac{0.3n_0}{n_0}) = -\frac{mgy}{KbT}

g = -\frac{KbT ln(0.3)}{my}

g = -\frac{(1.38*10^{-23}J/K)(20K)(Ln(0.3))}{10^{-19}(3*10^{-2})}

g = \frac{1.38*2*ln(0.3)*10^{-22}}{3*10^{-4}}

g = 1.104*10^{-1}m/s^2

g = 0.1m/s^2

Therefore the correct answer is C.

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4 years ago
can someone help me answer this? i dont get it and I need help before the next 30 minutes. ill give a brainiest to whoever helps
Zinaida [17]

Answer:

S=5.83...

Explanation:

s=d/t

Speed=distance/time

s=35/10

s=5.83...

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What type of tissue in the heart pumps blood throughout the body?
slavikrds [6]

Answer:

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