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2 years ago

Low tide____.

Chemistry

2 answers:

Fantom [35]2 years ago

8 0

Answer:

i think the answer is A

Explanation:

because in some areas, a regular pattern occurs of one high tide and one low tide each day,

marin [14]2 years ago

6 0

Answer:

A) happens once a day.

Explanation:

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What is the name of the compound made from Sn 4+ and PO4 3-?

givi [52]

Answer:

Sn₃(PO₄)₄ - tin(IV) phosphate.

Explanation:

Hope it helps! :)

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2 years ago

Terry, a student, performs a titration. He completes these steps as part of his titration procedure: 1. He cleans and rinses a b

Contact [7]

Answer:

Explanation:

D is the answer because 2 is needed to know

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2 years ago

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The acetic acid/acetate buffer system is a common buffer used in the laboratory. Write the equilibrium equation for the acetic a

hichkok12 [17]

Answer:

CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base or vice versa.

For the acetic acid buffer, CH₃CO₂H is the weak acid and its conjugate base is the ion without H⁺, that is CH₃CO₂⁻. The equilibrium equation in water knowing this is:

<em>In the equilibrium, the acid is dissociated in the conjugate base and the hydronium ion.</em>

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2 years ago

A student dissolves 73 g of sodium nitrate, NaNO3, in 100 mL of water and observes a clear, colorless

ohaa [14]

I 90% sure that it would be supersaturated

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2 years ago

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

2 years ago