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2 years ago

What drug is the greatest challenge for Forensic Science? What drug in specific is a threat in Forensics.

2 answers:

5 0

The greatest challenge to forensic science isn't any one drug, but a combination (mixture in an appropriate way) of drugs and chemicals and their complex reactions. The longer the drug remains in the body, the more it is metabolized into another substance.

[Letter is bolds, are important concept here ]

Hope this helps!

kati45 [8]2 years ago

3 0

Marijuana is a key drug in Forensic Science

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According to the principle of the conservation of mass the mass of the sodium sulfate was the following PLS HELP I NEED TO TURN

Annette [7]

Answer:

Explanation:

The Law of Conservation of Mass is defined and explained using examples of reacting mass calculations using the law are fully explained with worked out examples using the balanced symbol equation. The method involves reacting masses deduced from the balanced symbol equation.

5 0

2 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

3 years ago

What is the pH of a KOH solution that has [H ] = 1. 87 × 10–13 M? What is the pOH of a KOH solution that has [OH− ] = 5. 81 × 10

vlada-n [284]

pH is the hydrogen ion concentration and pOH is the hydroxide ion concentration in the solution. pH KOH is 12.73, pOH KOH is 2.24 and pH NaCl is 7.

pH is the negative log of the hydrogen ion concentration and pOH is the negative log of the hydroxide ion concentration.

The relation between the pH and pOH can be given as, $\rm pOH = 14 - pH$

The pH of KOH can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the first case, the concentration of the KOH is $1. 87 \times 10^{-13}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 87 \times 10^{-13}\;\rm M ]\\\\&= 12.73\end{aligned}$

Hence, the pH of KOH is 12.73.

pOH of KOH can be calculated by the formula,

$\rm pOH = \rm -log [OH^{-}]$

The hydroxide concentration of the KOH solution is $5. 81 \times 10^{-3}\;\rm M$

Substituting value in the equation:

$\begin{aligned} \rm pOH &= \rm -log [OH^{-}]\\\\&= \rm -log [5. 81 \times 10^{-3}\;\rm M ]\\\\&= 2.24 \end{aligned}$

Hence, the pOH of KOH is 2.24

The pH of NaCl can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the third case, the concentration of the NaCl is $1. 00\times 10^{-7}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 00 \times 10^{-7}\;\rm M ]\\\\&= 7 \end{aligned}$

Hence, the pH of KOH is 7.0.

Therefore, KOH is basic and NaCl is approximately neutral.

Learn more about pH and pOH here:

brainly.com/question/13885794

3 0

2 years ago

Silver was precipitated and collected by filtration. The experiment yielded .077g of silver after dying. The predicted yield was

PSYCHO15rus [73]

Answer:

Precentage error = 7.23%

Explanation:

The precentage error formula is the useful method for determining precision of your experimental result

It can be calculated using :

$percent\ error = |\frac{experimental\ value-predicted\ value} {predicted\ value}|\times 100$

$percent\ error = |\frac{0.077\ -0.083} {0.083}|\times 100$

So we get percentage error = 7.23%

Precision means how close the experimental value to the true value or theoretical value

If percentage error is large it means experimental results are deviated and there is need to check mistakes,imprecision in the experiment

If the percentage error is zero it means the experiment is perfectly done without inaccuracy

4 0

3 years ago

A sample of gas has a volume of 100. L at 17 °C and 800. torr. To what temperature must the gas be cooled in order for its volum

Romashka [77]

Answer:

108.81 K

Explanation:

First convert 17 °C to Kelvin:

17 + 273.16 = 290.16 K

Assuming ideal behaviour, we can solve this problem by using the combined gas law, which states that at constant composition:

P₁V₁T₂=P₂V₂T₁

Where in this case:

P₁ = 800 torr

V₁ = 100 L

T₂ = ?

P₂ = 600 torr

V₂ = 50 L

T₁ = 290.16 K

We input the data:

800 torr * 100 L * T₂ = 600 torr * 50 L * 290.16 K

And solve for T₂:

T₂ = 108.81 K

6 0

2 years ago