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lakkis [162]
3 years ago
9

Where do nutrients(food and water) enter the body first?

Chemistry
2 answers:
Gnom [1K]3 years ago
5 0

Answer:

esophagus……………..

klasskru [66]3 years ago
4 0
The mouth.
You first put food into your mouth
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The test tube that contained the calcium metal was probably the most difficult of the test tubes to clean. Why was aqueous HCl u
natita [175]

Answer:

The HCl will dissolve the calcium readily

Explanation:

HCl will react with d calcium, dissolving it to produce Hydrogen gas

6 0
3 years ago
Help please i have 5 minutes to do this !!!
kramer

Answer:

A) One that occurs on its own

4 0
3 years ago
Water is 11% hydrogen and 89% oxygen by mass. How many grams of oxygen are in a 250 g glass of water?
mylen [45]

Answer:

mass O2 = 222.5 g

Explanation:

  • %wt = ((mass compound)/(mass sln))×100

balance reaction:

  • 2H2 + O2 ↔ 2H2O

∴ %wt H2 = 11 % = ((mass H2)/(mass H2O))×100

∴ %wt O2 = 89 % = ((mass O2)/(mass H2O))×100

∴ mass H2O 250 g

⇒ mass O2 = (0.89)(250 g)

⇒ mass O2 = 222.5 g

3 0
3 years ago
This section of the periodic table is called a(n)
Mice21 [21]

Answer:

Is it <u>Group</u>?

Explanation:

6 0
3 years ago
Read 2 more answers
The balanced equation for water is 2 H2 + O2 to 2 H2O. If I have 21.2g of a product , and I started with 5.6 g of H2, how many g
ANTONII [103]

Since 21.2 g H2O was produced, the amount of oxygen that reacted can be obtained using stoichiometry. The balanced equation was given:  2H₂ + O₂ → 2H₂O and the molar masses of the relevant species are also listed below. Thus, the following equation is used to determine the amount of oxygen consumed.

Molar mass of H2O = 18 g/mol

Molar mass of O2 = 32 g/mol

21.2 g H20 x 1 mol H2O/ 18 g H2O x 1 mol O2/ 2 mol H2O x 32 g O2/ 1 mol O2 = 18.8444 g O2

<span>We then determine that 18.84 g of O2 reacted to form 21.2 g H2O based on stoichiometry. It is important to note that we do not need to consider the amount of H2 since we can derive the amount of O2 from the product. Additionally, the amount of H2 is in excess in the reaction.</span>

7 0
4 years ago
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