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dangina [55]
2 years ago
12

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

Chemistry
1 answer:
Rasek [7]2 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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Molecules can be composed of ____. Select all that apply. … two of more of the same atoms two or more smaller molecules two or m
sertanlavr [38]

Answer:

Explanation:

A group of two or more smaller molecules is the correct answer

3 0
2 years ago
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alekssr [168]
The answer is c Yep Allll day
6 0
2 years ago
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What is the mass of 8.12 × 10^23 molecules of CO2 gas? (Atomic mass of carbon = 12.011 u; oxygen = 15.999 u.)
amid [387]

Answer:

m=59.3gCO_2

Explanation:

Hello,

In this case, the first step is to compute the molar mass of carbon dioxide as shown below, considering it has one carbon atom and two oxygen atoms:

M=12.011g/mol+2*15.999g/mol\\\\M=44.009g/mol

It is important to notice it is the mass in one mole of such compound. Afterwards, we need to use the Avogadro's number to compute the how many moles are in the given molecules of carbon dioxide as shown below:

mol=8.12x10^{23}molec*\frac{1mol}{6.022x10^{23}molec} =1.35mol

Finally, the mass by using the molar mass:

m=1.35mol*\frac{44.009g}{1mol} \\\\m=59.3gCO_2

Best regards.

4 0
2 years ago
Your favorite beach has 42 large sand dunes throughout the course of the year wind erosion destroys eight sand dunes and create
a_sh-v [17]
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8 0
3 years ago
What is the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen
natta225 [31]
Carbon(C):
number of moles= mass/molar mass(Mr)
=65.5/12
=5.5 moles

Hydrogen(H):
number of moles=mass/molar mass (Mr)
=5.5/1
=5.5 moles

Oxygen (O):
number of moles = mass/molar mass (Mr)
=29.0/16
=1.8 moles

EF= lowest number of moles over each of the elements

So,
C= 5.5/1.8 = 3
H= 5.5/1.8 = 3
O= 1.8/1.8 = 1

Therefore Emperical formula= C3H3O
6 0
2 years ago
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