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algol13
2 years ago
9

Round the mmHg readings to the nearest whole number. Round the kPa readings to the nearest tenth. Round the atm readings to the

nearest hundredth. Convert a pressure of 1.76 atm to _____ torr.
Chemistry
1 answer:
Eva8 [605]2 years ago
6 0

i love jesus and i love my mommy

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For lunch today, I ate an apple. What type of carbohydrate did I ingest?
mafiozo [28]

Answer:

Monosaccharide

Explanation:

Apples contain high levels of fructose, which is a monosaccharide.

8 0
3 years ago
________________ are a lot like gases, but the atoms are different because they are made up of free electrons and ions of the el
lianna [129]

Answer:

plasma

Explanation:

Plasmas are a lot like gases, but the atoms are different, because they are made up of free electrons and ions of an element such as neon (Ne). You don't find naturally occurring plasmas too often when you walk around. They aren't things that happen regularly on Earth.

3 0
3 years ago
This reaction supports the Law of Conservation of Mass (hint: is the equation balanced?):
enyata [817]

Answer:

False

Explanation:

<em>Left:</em>

K = 2 atoms

O = 1 atoms

H = 1 atoms

<em>Right: </em>

K = 2 atoms

O = 1 atoms

H = 2 atoms

O = 1 atoms

As you can see both sides do not have the same amount of atoms, therefore it is not balanced.

<h3>Hope this Helps!!</h3>

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7 0
3 years ago
AgCl is..<br> A)an element <br> B)a compound <br> C.)a mixture
SOVA2 [1]
It is an A compound
6 0
2 years ago
Read 2 more answers
Propane (C3H8) can be burned to produce heat for homes. The products of the reaction are CO2 and H2O. For complete combustion to
Natalija [7]

Answer:

1- 3 Moles of CO2  

2- 132 g of CO2  

3- 105,6 g of CO2

4- Limiting Reagent O2

<u>Products form based on limiting reagent (384g O2) :</u>

CO2: 316,8 g

H2O: 172,8 g

<u>Products form based on C3H8 (132,33 g):</u>

CO2: 396,99 g

H2O: 216,54 g  

Explanation:

<u>Atomic Masses:</u>

C: 12

H: 1

O: 16

<u>Molecular weights:</u>

C3H8: 44 g

O2: 32 g

H2O: 18 g

CO2: 44 g

C3H8 + 5 O2⇒ 3 CO2 + 4 H2O

C3H8 (44g)+ O2 (160 g) ⇒ CO2 (132 g) + H2O (72 g)

In 5 moles of O2 are produced 3 moles of CO2, equivalent to 132 g

For 160g of O2 are produced 132 g of CO2, so 128 g of O2

160 g  O2 ⇒ 132 g CO2

128 g  O2⇒ × = 105,6 g CO2

(128×132÷160= 105,6)

The limiting reagent is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, because the reaction cannot continue without it.

If I have 132,33 g of C3H8 and 384 g O2 we can calculate:

For          44 g of CH3H8  ⇒160 g of O2

With   132,33 g of CH3H8 ⇒ ×= 481,2 g of O2

(132,33×160÷44=481,2)

As this amount exceeds the quantity of O2 that we have, we can assume that the 384 g O2 will be totally consumed.

<u>Calculations of the products formed in base of quantity of O2 (limiting reagent):</u>

160 g  O2 ⇒ 132 g CO2

384 g  O2⇒ × = 316,8 g CO2

(384×132÷160= 316,8)

160 g  O2 ⇒ 72 g H2O

384 g  O2⇒ × = 172,8 g H2O

(384×72÷160= 172,8)

<u>Calculations of the products formed in base of quantity of C3H8 (excess reagent):</u>

     44 g  C3H8 ⇒ 132 g CO2

132,33 g  C3H8 ⇒ × = 396,99 g CO2

(132,33×132÷44=396,99)

     44 g C3H8 ⇒ 72 g H2O

132,33 g  C3H8⇒ × = 216,54 g H2O

(132,33×72÷44= 216,54)

<u />

3 0
3 years ago
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