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kotykmax [81]
3 years ago
7

Which of the following could be considered a scientific statement?

Chemistry
2 answers:
schepotkina [342]3 years ago
8 0
Ants live In colonies
Alexus [3.1K]3 years ago
6 0

Answer: Ants live in colonies.

Explanation:

The 1st, 3rd, and 4th, answer choice are all considered as opinions.

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Calculate the formula for the following hydrate composed of 76.9% CaSO3 and 23.1%H2O
artcher [175]
Assume there is 100g of the substance at first

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In a chemical change
Vsevolod [243]

Answer:

B. materials change their properties.

Explanation:

In a chemical change, materials often change their properties because a re-arrangement of atoms takes place.

A chemical change is one in which new kind of matter is formed.

It is always accompanied by energy changes.

  • Chemical changes are not reversible.
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3 years ago
How much sucrose (g) do you need to weight in order to prepare 19.16 g of a 13.1 % (weight percent) solution?
Nonamiya [84]
<span>2.51 grams
   You want to prepare 19.16 g of some solution which will have 13.1% of it's mass being sucrose. So we just need to perform some simple multiplication: 19.16g * 0.131 = 2.50996g
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3 0
3 years ago
How many moles of CO2 are emitted into the atmosphere when 22.1 g C8H18 is burned
ANTONII [103]

Answer:

1.552 moles

Explanation:

First, we'll begin by writing a balanced equation for the reaction showing how C8H18 is burn in air to produce CO2.

This is illustrated below:

2C8H18 + 25O2 -> 16CO2 + 18H2O

Next, let us calculate the number of mole of C8H18 present in 22.1g of C8H18. This is illustrated below:

Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol

Mass of C8H18 = 22.1g

Mole of C8H18 =..?

Number of mole = Mass /Molar Mass

Mole of C8H18 = 22.1/144

Mole of C8H18 = 0.194 mole

From the balanced equation above,

2 moles of C8H18 produced 16 moles of CO2.

Therefore, 0.194 mole of C8H18 will produce = (0.194x16)/2 = 1.552 moles of CO2.

Therefore, 1.552 moles of CO2 are emitted into the atmosphere when 22.1 g C8H18 is burned

8 0
3 years ago
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