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dangina [55]
3 years ago
12

3- AR 385-63/MCO 3570.1C is used in conjunction with

Chemistry
1 answer:
Rasek [7]3 years ago
8 0
AR 385-63/MCO 3570.1C is used in conjunction with DA PAM 385-63.
AR 385-63/MCO 3570.1C is the regulation or order which provides revised range safety policy for the Army and Marine Corps.
This order/regulation applies on the Active Army or the Army National Guard of the United states.
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A city's water supply is contaminated with a toxin at a concentration of 0.63 mg/L. For the water to be safe for drinking, the c
Shalnov [3]

Answer:

Approximately 22.37 days, will it take for the water to be safe to drink.

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

k is rate constant

Given that:- k = 0.27 (day)⁻¹

[A_0] = 0.63 mg/L

[A_t]=1.5\times 10^{-3} mg/L

Applying in the above equation as:-

1.5\times 10^{-3}=0.63e^{-0.27\times t}

63e^{-0.27t}=150\times \:10^{-3}

e^{-0.27t}=\frac{1}{420}

t=\frac{100\ln \left(420\right)}{27}=22.37

<u>Approximately 22.37 days, will it take for the water to be safe to drink.</u>

7 0
3 years ago
Why do we call meteors shooting stars?
zlopas [31]

They burn up and cause it to look like shooting stars

6 0
3 years ago
How many sulfur atoms are present in 25.6 g of Al2(S2O3)3
IgorC [24]

Given the mass of Al_{2}(S_{2}O_{3})_{3}=25.6 g

The molar mass of Al_{2}(S_{2}O_{3})_{3}=390.35g/mol

Converting mass of Al_{2}(S_{2}O_{3})_{3}to moles:

25.6 g Al_{2}(S_{2}O_{3})_{3}*\frac{1molAl_{2}(S_{2}O_{3}}{390.35 gAl_{2}(S_{2}O_{3}} =0.0656molAl_{2}(S_{2}O_{3}

Converting mol Al_{2}(S_{2}O_{3})_{3}to mol S:

0.0656mol Al_{2}(S_{2}O_{3})_{3}*\frac{6molS}{1mol Al_{2}(S_{2}O_{3})_{3}}=0.3936 molS

Converting mol S to atoms of S using Avogadro's number:

1 mol = 6.022*10^{23}atoms

0.3936mol S *\frac{6.022*10^{23}atoms S}{1 mol S}=2.37*10^{23} S atoms

5 0
3 years ago
Plants and animals add carbon dioxide to the atomsphere when they ____ glucose.
labwork [276]
A! Oxidize. Hope this helps!
7 0
3 years ago
If 73.5 mL of 0.200 M KI(aq) was required to precipitate all of the lead(II) ion from an aqueous solution of lead(II) nitrate, h
frutty [35]

Answer:

There were 0.00735 moles Pb^2+ in the solution

Explanation:

Step 1: Data given

Volume of the KI solution = 73.5 mL = 0.0735 L

Molarity of the KI solution = 0.200 M

Step 2: The balanced equation

2KI + Pb2+ → PbI2 + 2K+

Step 3: Calculate moles KI

moles = Molarity * volume

moles KI = 0.200M * 0.0735L = 0.0147 moles KI

Ste p 4: Calculate moles Pb^2+

For 2 moles KI we need 1 mol Pb^2+ to produce 1 mol PbI2 and 2 moles K+

For 0.0147 moles KI we need 0.0147 / 2 = 0.00735 moles Pb^2+

There were 0.00735 moles Pb^2+ in the solution

3 0
3 years ago
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