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Sever21 [200]
3 years ago
8

What is concentration

Chemistry
2 answers:
Sauron [17]3 years ago
8 0

Answer:

the action or power of focusing one's attention or mental effort.

Explanation:

Slav-nsk [51]3 years ago
3 0

In chemistry, concentration is the abundance of a constituent divided by the total volume of a mixture.

Explanation:

if non science related concentration is also about focusing(non-science related definition) : )

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When selling on the street, dealers may not know the purity of the ketamine they have, and thus users do not know exactly how mu
fiasKO [112]

Answer:

a. 6.15 mL b. 30.73 mL

Explanation:

a. What volume of this ketamine solution would the 65.0 kg user have to inject to experience a high at 0.400 mg/kg?

Since we have 0.250 g of ketamine in 1/4 cup of water and 1 cup of water equals 236.5 mL, we need to find the concentration of ketamine we have.

So concentration of ketamine C = mass of ketamine, m/volume of water, V

m = 0.250 g and V = 1/4 cup = 1/4 × 236.5 mL = 59.125 mL

So, C = m/V = 0.250 g/59.125 mL = 0.00423 g/mL = 4.23 mg/mL

Since the user has a mass of 65 kg and requires a high at 0.400 mg/kg, the mass of ketamine for this high is M = 65 kg × 0.400 mg/kg = 26 mg

Since mass, M = concentration ,C × volume, V

M = CV

V = M/C

The volume of ketamine required for the 0.400 mg/kg high is

V = 26 mg/4.23 mg/mL

V = 6.15 mL

b. What volume of this ketamine solution would the user have to inject to become unconscious at 2.00 mg/kg?

Since the concentration of ketamine is C = 4.23 mg/mL, and Since the user has a mass of 65 kg and requires an injection of 2.00 mg/kg to be unconscious, the mass of ketamine required to be unconscious is M' = 65 kg × 2.00 mg/kg = 130 mg

Since mass, M' = concentration ,C × volume, V

M' = CV

V = M/C

The volume of ketamine required for the 2.00 mg/kg unconscious injection is

V = 130 mg/4.23 mg/mL

V = 30.73 mL

5 0
2 years ago
What descriptive term is applied to the type of diene represented by 2,4-hexadiene?.
Varvara68 [4.7K]

The descriptive term applied to the type of diene represented by 2,4-hexadiene is conjugated diene.

Dienes are compounds which contains two double bonds. These dienes can be non conjugated or conjugated.

Conjugated diene are those compound which have two double bonds  joined by a single σ bond. Conjugated dienes can also be called 1,3-diene. To know if diene is conjugated or non conjugated, sp³ hybridization is to b checked and the number of double bonds and single sigma bond is checked.

Conjugated dienes are found in many different molecules. 2,4-hexadiene is a conjugated diene with two carbon-carbon double bonds that are separated by one sigma bond.

The stabilization of dienes by conjugation is better than the aromatic stabilization. Conjugated dienes are more stable than non conjugated or cumulative diene because it has higher electron density of molecules delocalized.

To learn more about conjugated dienes,

brainly.com/question/24261651

#SPJ4

6 0
1 year ago
How many protons are in an element with an atomic number of 8 and a mass number of 18?
Natalija [7]
The answer is 8. atomic number: the number of protons in the nucleus of an atom.
4 0
3 years ago
Read 2 more answers
If an object has a mass of 250 grams and a volume of 5 milliliters, what is the density of the object?
ycow [4]

Answer:

<h2>50 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question we have

density =  \frac{250}{5}  \\

We have the final answer as

<h3>50 g/mL</h3>

Hope this helps you

4 0
2 years ago
Tris is a molecule that can be used to prepare buffers for biochemical experiments. It exists in two forms: Tris (a base) and Tr
ale4655 [162]

Solution :

For the reaction :

$\text{TrisH}^+ + H_2O \rightarrow \text{Trish}^- + H_3O^+$

we have

$Ka = \frac{[\text{Tris}^- \times H_3O]}{\text{Tris}^+}$

   $=\frac{x^2}{0.02 -x}$

  $=8.32 \times 10^{-9}$

Clearing $x$, we have $x = 1.29 \times 10^{-5} \text{ moles of acid}$

So to reach $\text{pH} = 7.8 (\text{pOH}= 14-7.8=6.2)$, one must have the $\text{OH}^-$ concentration of the :

$\text{[OH}^-]=10^{-pOH} = 6.31 \times 10^{-7} \text{ moles of base}$

So we can add enough of 1 M NaOH in order to neutralize the acid that is calculated above and also adding the calculated base.

$\text {n NaOH}=1.29 \times 10^{-5}+6.31 \times 10^{-7}$

            $= 1.35 \times 10^{-5} \text{ moles}$

Volume NaOH $= 1.35 \times 10^{-5} \text{ moles} \times \frac{1000 \text{ mL}}{1 \text{ mol}} = 0.0135 \text{ mL}$

Tris mass $H^+ = 0.02 \text{ mol} \times 157.6 \text{ g/mol}=3.152 \text{ g}$

Now to prepare the said solution we must mix:

$3.152 \text{ g Tris H} + 0.0135 \text{ mL NaOH} \ 1 M$ gauge to 1000 mL with water.

3 0
3 years ago
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