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Natali5045456 [20]
3 years ago
13

Ethylene glycol (antifreeze) has a density of 1.11 g/cm3.

Chemistry
1 answer:
disa [49]3 years ago
7 0
Hope this helps you.

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E=mc^2 and mass defect
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PLEASE HELP ME URGENT!!! How do i find mass when not given in chemistry
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It depends what other values you have. Can you give more info? If they give density then you can solve for m.

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Sodium hydrogen carbonate and citric acid<br><br> Word equation
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the right answer is:

Explanation:

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What type of cleaning agent should be used to clean a laminar flow hood?
lara31 [8.8K]

The type of cleaning agent that should be used to clean a laminar flow hood is 70% Isopropyl Alcohol.

Isopropyl alcohol is mixed with water for use as a rubbing-alcohol antiseptic. It is also utilized in aftershave creams, hand lotions, and other cosmetics. In enterprise, it is used as a less expensive solvent for cosmetics, drugs, shellacs, and gums, in addition to denaturing ethanol (ethyl alcohol).

Isopropyl alcohol is a colorless, flammable chemical compound with a robust alcoholic smell. As an isopropyl institution related to a hydroxyl group, it's far the handiest example of a secondary alcohol, wherein the alcohol carbon atom is connected to 2 other carbon atoms.

Isopropyl alcohol is pure alcohol and is a colorless liquid with a musty, sharp odor. There are no other components in a bottle of isopropyl alcohol. through assessment, rubbing alcohol consists of isopropyl alcohol among different components, including water. maximum rubbing alcohol manufacturers incorporate 70% isopropyl alcohol.

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#SPJ4

6 0
2 years ago
If a graduated cylinder has 28ml of water in it and you add an object which causes the water level to rise to 47ml, what is the
grin007 [14]

Answer:

The object placed in the water has a volume of 19 cm³

Explanation:

<u>Step 1: </u>Data given

volume of the cylinder before adding the object = 28 mL = 28 cm³

After adding an object with volume X the volume rises to 47 mL = 47 cm³

<u>Step 2:</u> Calculate the volume of the object

Volume of the object = Final volume - initial volume

Volume of the object = 47 cm³ (or 47 mL) - 28 cm³ ( or 28 mL) = 19 cm³ (or 19 mL)

The object placed in the water has a volume of 19 cm³

7 0
3 years ago
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