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3 years ago

Question 1 (1 point)

Chemistry

1 answer:

Colt1911 [192]3 years ago

8 0

Answer:

Both are made up of two substances that are not chemically combined.
oxygen (O2)
carbon dioxide (CO2 )
air (N2 mixed with O2 and CO2)
Pure substances cannot be separated by physical means.

Explanation:

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How do balanced chemical equations show the conservation of mass

antiseptic1488 [7]

Answer: Matter cannot be created or destroyed

Explanation: Balanced equations are set equations we cannot change one element or compound in the equation without changing the entire equation. So balanced equation show the conservation of mass because while other substances may be formed from the synthesis or decomposition of compounds new elements are never introduced and are not created out of thin air :)

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3 years ago

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an ionic compound is typically made of a. two metals. b. two nonmetals. c. a metal and a nonmetal. d. none of the above.

kirill115 [55]

C. a metal and a nonmetal

6 0

3 years ago

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How many grams are in 1.76 x 10^23 atoms of iodine

Mariana [72]

Answer:

$\boxed {\boxed {\sf About \ 37.1 \ grams \ of \ iodine }}$

Explanation:

To convert from atoms to grams, you must first convert atoms to moles, then moles to grams.

1. Convert Atoms to Moles

To convert atoms to grams, Avogadro's number must be used.

$6.022*10^{23}$

This number tells us the number of particles (atoms, molecules, ions, etc.) in 1 mole. In this case, the particles are atoms of iodine.

$\frac{6.022*10^{23} \ atoms \ I }{1 \ mol \ I}$

Multiply the given number of atoms by Avogadro's number.

$1.76*10^{23} \ atoms \ I*\frac{6.022*10^{23} \ atoms \ I }{1 \ mol \ I}$

Flip the fraction so the atoms of iodine will cancel out.

$1.76*10^{23} \ atoms \ I*\frac{ 1 \ mol \ I}{6.022*10^{23} \ atoms \ I}$

$1.76*10^{23}* \frac{1 \ mol \ I}{6.022*10^{23} }$

Multiply so the problem condenses into 1 fraction.

$\frac{1.76*10^{23} \ mol \ I}{6.022*10^{23} }$

$0.2922617071 \ mol \ I$

2. Convert Moles to Grams

Now we must use the molar mass of iodine, which is found on the Periodic Table.

Iodine Molar Mass: 126.9045 g/mol

Use this mass as a fraction.

$\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}$

Multiply this fraction by the number of moles found above.

$0.2922617071 \ mol \ I*\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}$

Multiply. The moles of iodine will cancel.

$0.2922617071 *\frac{ 126.9045 \ g\ I }{ 1 }$

The 1 as a denominator is insignificant.

$0.2922617071 *{ 126.9045 \ g\ I }$

$37.08932581 \ g \ I$

3. Round

The original measurement of 1.76*10^23 has 3 significant figures (1, 7, and 6). Therefore we must round our answer to 3 sig figs. For this answer, that is the tenths place.

$37.08932581 \ g \ I$

The 8 in the hundredth place tells us to round the 0 up to a 1.

$\approx 37.1\ g \ I$

There is about <u>37.1 grams of iodine </u>in 1.76*10^23 atoms.

5 0

3 years ago

Please help

Alika [10]

Answer:

there is no question to answer :(

4 0

3 years ago

a substance is analyzed to have a percent composition of 74.186% sodium and 25.814% fluorine. calculate the empirical formula

Novosadov [1.4K]

Answer:

Na₂₆F₁₁

Explanation:

We find the moles of the substance assuming 100 g of the substance is present. Why do we take 100 g? Because then the percent of sodium/fluorine, would be the g of sodium/fluorine respectively:

74.186 g Sodium | 1 mol Sodium/23 g => 3.2255 mol Na

25.814 g Fluorine | 1 mol Fluorine/19 g => 1.3586 mol F

Divide each by smallest number of moles:

3.2255/1.3586 = 2.37

1.3586/1.3586 = 1

Multiply by common number to get a smallest whole number:

2.37*11 = 26,

1*11 = 11

The empirical formula is Na₂₆F₁₁

5 0

3 years ago