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inessss [21]
3 years ago
13

Self meditating to deal with stress

Chemistry
1 answer:
hjlf3 years ago
3 0

I think the answer is C but I also feel it should be A...... but I think its C

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A mole of ethyl alcohol weighs 46 g. how many grams of ethyl alcohol are needed to produce 1 l of a 2-millimolar (2 mm) solution
Thepotemich [5.8K]
Answer: 92mg of ethyl alcohol 

2 milli molar of ethyl alcohol means 2 milli mole glucose for 1 liter of water. Since we want to make 1L of solution, then the amount of ethyl alcohol needed is: 2 milli mole/l x 1l= = 2 x 10^-3 mole

Ethyl alcohol molecular weight is 46g/mole, then 2 mili mole of glucose is= 2x 10^-3 mole x 46 grams/mole= 92 grams x 10^-3= 92mg
7 0
4 years ago
The neuron functions to
patriot [66]

Answer:

B) neuron or nerve cell work is to carry messages

7 0
3 years ago
Equal volumes of 0.2 M ammonia and 0.2 M nitric acid, HNO₃, are mixed. Write the net ionic equation for the reaction and identif
Black_prince [1.1K]

Answer:

The net ionic equation is

H₃O⁺+NH₃ ↔ NH₄⁺+H₂O

Explanation:

To write the net ionic equation, we are required to dissociate the into ions all strong acids and strong bases

Hence, nitric acid which is a strong acid is dissociated as follows

HNO₃+H₂O → H₃O⁺ + NO₃⁻

in the above equation, the nitrate ion  NO₃⁻, is a spectator ion because it is only present and does not partake in the chemical reaction so it is left out of the net ionic equation equation

Also the it is required to keep together weak bases in the solution therefore  for NH₃ which is a weak base we have

NH₃ + H₃O⁺ → NH₄⁺ + H₂O

Hence, the net ionic equation becomes

H₃O⁺ (aq) + NH₃ ↔ NH₄⁺(aq) + H₂O (l)

3 0
4 years ago
What is the wave speed <br>a wave with a wave length of 1000 m and a frequancy of 3000000 Hz ​
Alla [95]

Answer:

v = 3,000,000,000 m/s

Explanation:

v = f *L

v = 3,000,000 * 1000

v = 3,000,000,000 m/s

7 0
3 years ago
Calculate the molar concentrations of H+ and OH− in solutions that have the following pH values:
Anastaziya [24]

Answer:

Explanation:

a )

pH = - log[ H⁺]

8.26 = - log[ H⁺]

[ H⁺] = 10⁻⁸°²⁶ mole / l

= 5.49 x 10⁻⁹ moles / l

[ H⁺] [OH⁻] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 5.49 x 10⁻⁹

= .182  x 10⁻⁵ moles / l

b )

10.25 = - log[ H⁺]

[ H⁺] = 10⁻¹⁰°²⁵ mole / l

= 5.62 x 10⁻¹¹ moles / l

[ H⁺] [OH⁻] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 5.62 x 10⁻¹¹

= .178 x 10⁻³ moles / l

c )

4.65 = - log[ H⁺]

[ H⁺] = 10⁻⁴°⁶⁵ mole / l

= 2.24 x 10⁻⁵ moles / l

[ H⁺] [OH⁻] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 2.24 x 10⁻⁵

= .4464 x 10⁻⁹ moles / l

7 0
3 years ago
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