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Alex Ar [27]
2 years ago
6

How many moles of carbonate are there in sodium carbonate​

Chemistry
2 answers:
vaieri [72.5K]2 years ago
5 0

There are 0.566 moles of carbonate in sodium carbonate.

<h3>CALCULATE MOLES:</h3>
  • The number of moles of carbonate (CO3) in sodium carbonate (Na2CO3) can be calculated by dividing the mass of carbonate in the compound by the molar mass of the compound.

  • no. of moles of CO3 = mass of CO3 ÷ molar mass of Na2CO3

  • Molar mass of Na2CO3 = 23(2) + 12 + 16(3)

  • = 46 + 12 + 48 = 106g/mol

  • mass of CO3 = 12 + 48 = 60g

  • no. of moles of CO3 = 60/106

  • no. of moles of CO3 = 0.566mol

  • Therefore, there are 0.566 moles of carbonate in sodium carbonate.

Learn more about number of moles at: brainly.com/question/1542846

ivolga24 [154]2 years ago
4 0
<h2>Answer:</h2>

molecular weight of Sodium Carbonate or mol The molecular formula for Sodium Carbonate is Na2CO3. The SI base unit for amount of substance is the mole.','.

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A student made a graph to show the chemical equilibrium position of a reaction.
pogonyaev

Answer:

Concentration, because the amounts of reactants and products remain constant after equilibrium is reached.

Explanation:

The rate of reaction refers to the amount of reactants converted or products formed per unit time.

As the reaction progresses, reactions are converted into products. This continues until equilibrium is attained in a closed system.

When equilibrium is attained, the rate of forward reaction is equal to the rate of reverse reaction, hence the concentration of reactants and products in the system remain fairly constant over time.

When deducing the rate of reaction, concentration of the specie of interest is plotted on the y-axis against time on the x-axis.

8 0
3 years ago
What is the mass of 4.99×1021 platinum atoms?
charle [14.2K]

Answer:

\boxed {\boxed {\sf 1.62 \ g \ Pt}}

Explanation:

We are asked to find the mass of a number of platinum (Pt) atoms.

<h3>1. Convert Atoms to Moles </h3>

First, we convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are platinum atoms.

We will convert using dimensional analysis so we set up a ratio using the information we know (6.022 × 10²³ platinum atoms in 1 mole of platinum).

\frac {6.022 \times 10^{23} \ atoms \ Pt}{ 1 \ mol \ Pt}

We are converting 4.99 ×10²¹ atoms of Pt to moles of Pt, so we multiply by this value.

4.99 \times 10^{21} \ atoms \ Pt *\frac {6.022 \times 10^{23} \ atoms \ Pt}{ 1 \ mol \ Pt}

Flip the ratio so the units of atoms of platinum cancel.

4.99 \times 10^{21} \ atoms \ Pt *\frac { 1 \ mol \ Pt}{6.022 \times 10^{23} \ atoms \ Pt}

4.99 \times 10^{21} *\frac { 1 \ mol \ Pt}{6.022 \times 10^{23}}

\frac { 4.99 \times 10^{21}}{6.022 \times 10^{23}} \ mol \ Pt

Divide.

0.008286283627 \ mol \ Pt

<h3>2. Convert Moles to Grams </h3>

Next, we convert moles to grams using the molar mass. This is the mass of 1 mole of a substance. This is found on the Periodic Table because it is equivalent to the atomic mass, but the units are grams per mole instead of atomic mass units. Look up platinum's molar mass.

  • Pt: 195.08 g/mol

Set up another ratio using this new information (195.08 grams of Pt in 1 mole of Pt).

\frac {195.08 \ g \ Pt}{ 1 \ mol \ Pt}

Multiply by the number of moles we just calculated.

0.008286283627 \ mol \ Pt*\frac {195.08 \ g \ Pt}{ 1 \ mol \ Pt}

The units of moles of platinum cancel.

0.008286283627*\frac {195.08 \ g \ Pt}{ 1 }

0.008286283627* {195.08 \ g \ Pt}

1.61648821\ g \ Pt

<h3>3. Round</h3>

The original measurement of atoms ( 4.99 ×10²¹ ) has 3 significant figures, so our answer must have the same. For the number we found, that is the hundredth place. The 6 in the thousandth place tells us to round the 1 up to a 2.

1.62 \ g \ Pt

There are approximately <u>1.62 grams of platinum</u> in 4.99 ×10²¹ atoms of platinum.

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In the following reaction, what element is gaining mass?
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6. What parts of a nuclear reactor keep the reaction from
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B. control rods and moderators

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2.  Swap the partners for all the other ions that you can get from step 1.  You can skip pairings with the same charge - a + can't get close to another + to react.
3.  Use solubility, acid/base, and redox rules to see if anything will happen with the ions in solution.<span />
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