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Andreyy89
3 years ago
15

To double-check a chemical in lab before using it, sniff near the opening of the container, true or false?

Chemistry
1 answer:
Bezzdna [24]3 years ago
5 0
The answers is false because you do not want to inhale unknown chemicals
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______ use energy from their enviornment to make food .
wlad13 [49]

Answer:

autotrophs

Explanation:

3 0
3 years ago
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You carefully weigh out 11.00 g of caco3 powder and add it to 44.55 g of hcl solution. you notice bubbles as a reaction takes pl
Zinaida [17]
The bubbles that were observed after the mixing of the two substances is one of the products of the reaction. It is the carbon dioxide that is produced. To determine the mass of this gas produced, we need to remember the Law of conservation of mass where mass cannot be created or destroyed. With this, we can say that the total mass that goes in a process should be equal to the mass that is goes out of the process no matter what the reaction is. We do as follows:

Mass of reactants = mass of products
11.00 + 44.55 = 51.04 + mass of carbon dioxide
mass of carbon dioxide = 4.51 g
8 0
4 years ago
Define and give 3 examples of psychical properties ​
Katen [24]

Answer:

A physical property is any property that is measurable, whose value describes a state of a physical system.

Explanation:

-color

-density

-volume

3 0
4 years ago
The Symbol below is used to represent what component in a circuit?
FrozenT [24]
Actually, I strongly believe it is a switch.
8 0
3 years ago
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A solution of hydrocloric acid has a molarity of 2.25 M HCl. If a reaction requires 5.80 g of HCI, what
Natali [406]

Answer:

0.071L

Explanation:

From the question given, we obtained the following data:

Molarity of HCl = 2.25 M

Mass of HCl = 5.80g

Molar Mass of HCl = 36.45g/mol

Number of mole of HCl =?

Number of mole = Mass /Molar Mass

Number of mole of HCl = 5.8/36.45 = 0.159mole

Now, we can obtain the volume required as follows:

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 0.159mole/ 2.25

Volume = 0.071L

4 0
4 years ago
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