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kirill115 [55]
3 years ago
6

Complete these metric conversions: 53 m - mm*

Chemistry
1 answer:
EleoNora [17]3 years ago
3 0

Answer:

53 meters = 53000 millimeters

Explanation:

In this question we have to convert meters into millimeters .

By metric conversion,

Since, 1 meter = 1000 mm

Therefore, 53 meters = 53 × 1000

                                  = 53000 millimeters

53 meters = 53000 millimeters is the answer.

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How many moles of sodium carbonate are contained by 57.3g of sodium carbonate
Lady_Fox [76]

Answer:

\boxed {\boxed {\sf 0.541 \  mol \ Na_2CO_3}}

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

  • Na:  22.9897693 g/mol
  • C: 12.011 g/mol
  • O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

  • Na₂ = 22.9897693 * 2= 45.9795386 g/mol
  • O₃ = 15.999 * 3= 47.997 g/mol
  • Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

We are converting 57.3 grams to moles, so we multiply by this value.

57.3 \ g \ Na_2CO_3} *\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

Flip the ratio so the units of grams of sodium carbonate cancel.

57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386  \ g \ Na_2CO_3}

57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }

\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3

0.5406295944 \ mol \ Na_2CO_3

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

0.541 \  mol \ Na_2CO_3

There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.

6 0
3 years ago
HELP ASAP!! FILL IN THE BLANKS! will be brainliest!!!
Ludmilka [50]

Answer:

Don't mark me brainliest because of this but I'm pretty sure your supposed to give us the words because teachers don't give you things like that without the words you will the answer in with.

4 0
3 years ago
Read 2 more answers
A molecule of common table salt, or nacl, is the result of _____ bond forming between a sodium (na) atom and a chlorine (cl) ato
jekas [21]
I believe the correct term to fill in the blank would be ionic. A molecule of common table salt, or nacl, is the result of ionic bond forming between a sodium (na) atom and a chlorine (cl) atom. Ionic bonding is a result of complete transfer of electrons between atoms. It usually happens between a metal and a nonmetal.
4 0
3 years ago
If the volume occupied by the air in a bicycle pump is 525 cm3, and the pressure changes from 73.2 kPa to 122.5 k.Pa as the pist
Musya8 [376]

Answer:

V_2=313.71\ cm^3

Explanation:

Given that,

Initial volume, V_1=525\ cm^3

The pressure changes from 73.2 kPa to 122.5 k.Pa.

We need to find the new volume occupied by the air. Let it is V₂. It can be calculated using Boyle's law such that,

P_1V_1=P_2V_2\\\\V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{73.2\times 525}{122.5}\\\\V_2=313.71\ cm^3

So, the new volume is 313.71\ cm^3.

6 0
3 years ago
The PH of a 0.1 M MCl (M is an unknown cation) was found to be 4.7. Write the net ionic equation for the hydrolysis of M and its
Serggg [28]

Answer:

6.25 X10^{-9} = Ka

Kb= \frac{10^{-14}}{6.25 X10^{-9}}= 1.6X10^{-6}

Explanation:

The ionic equation for the hydrolysis of the cation of the given salt will be:

M^{+} + H_{2}O ---> MOH + H^{+}

The expression for Ka will be:

Ka = \frac{[H^{+}][MOH]}{[M^{+}]}

As given that the concentration of the salt is 0.1 M and pH of solution = 4.7, we can determine the concentration of Hydrogen ions from the pH

pH = -log [H⁺]

[H⁺] = antilog(-pH) = antilog (-4.7) = 2 X 10⁻⁵ M = [MOH]

Let us calculate Ka from this,

Ka = {2.5X10^{-5}X2.5X10^{-5}}{0.1} = 6.25 X10^{-9}

The relation between Ka an Kb is

KaXKb =10⁻¹⁴

Kb= \frac{10^{-14}}{6.25 X10^{-9}}= 1.6X10^{-6}

6 0
3 years ago
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