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2 years ago

How many sodium ions are present in 325 ml of 0.850 m na2so4?

Chemistry

1 answer:

NISA [10]2 years ago

4 0

Answer:

$Na^+_{ions}=3.33x10^{23}Na^+_{ions}$

Explanation:

Hello,

In this case, for the given molarity and volume of such solution, the moles of sodium sulfate are computed below:

$n_{Na_2SO_4}=0.850\frac{mol}{L}*0.325L=0.276molNa_2SO_4$

Now, by using the Avogadro's number, the ions result:

$Na^+_{ions}=0.276molNa_2SO_4*\frac{2molNa^+}{1molNa_2SO_4} *\frac{6.022x10^{23}Na^+_{ions}}{1molNa^+} \\\\Na^+_{ions}=3.33x10^{23}Na^+_{ions}$

Best regards.

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Just Lemons Lemonade Recipe Equation:

zalisa [80]

Answer:

Explanation:

Hello!

Complete text:

Honors Stoichiometry Activity WorksheetInstructions:

Activity Two: Just Lemons, Inc. Production

Here's a one-batch sample of Just Lemons lemonade production. Determine the percent yield and amount of leftover ingredients for lemonade production and place your answers in the data chart.

Hint: Complete stoichiometry calculations for each ingredient to determine the theoretical yield. Complete a limiting reactant-to-excess reactant calculation for both excess ingredients.

Water 946.36 g

Sugar 196.86 g

Lemon Juice 193.37 g

Lemonade 2050.25g

Leftover Ingredients?

Just Lemons Lemonade Recipe Equation:

2 water + sugar + lemon juice = 4 lemonade

Mole conversion factors:

1 mole of water = 1 cup = 236.59 g

1 mole of sugar = 1 cup = 225 g

1 mole of lemon juice = 1 cup = 257.83 g

1 mole of lemonade = 1 cup = 719.42 g

You have the information on the ingredients used to produce one batch of lemonade and the amount of lemonade produced. To determine which ingredients be leftovers, you have to determine first, which one is the limiting reactant, i.e. the ingredient that will be used up first.

According to the recipe, to make 4 moles of lemonade, you use 2 moles of water, one mole of sugar and one mole of lemon juice, expressed in grams:

2 water + sugar + lemon juice = 4 lemonade

2*(236.59) + 225g + 257.83g = 4*(719.42)g

473.18g + 225g + 257.83g = 2877.68g

So for every 2877.68g of lemonade made, they use 473.18g of water, 225g of sugar, and 257.83g of lemon juice.

You know that they made a batch of 2050.25g, so to detect the limiting reactant, first, you have to calculate, in theory, how much of each ingredient you need to make the given amount of lemonade:

Use cross multiplication

Water:

2877.68g lemonade → 473.18g water

2050.25g lemonade → X= (2050.25*473.18)/2877.68= 337.12g water

Following the recipe, to elaborate 2050.25g of lemonade, you need to use 337.12g of water.

Sugar:

2877.68g lemonade → 225g sugar

2050.25g lemonade → X= (2050.25*225)/2877.68= 160.30g sugar.

To elaborate 2050.25f of lemonade you need to use 160.30g of sugar.

Lemon juice:

2877.68g lemonade → 257.83g lemon juice

2050.25g lemonade → X= (2050.25*257.83)/2877.68= 183.69g lemon juice.

To elaborate 2050.25f of lemonade you need to use 183.69g lemon juice.

Available ingredients vs. theoretical yields for 2050.25g of lemonade:

Water 946.36 g → 337.12g

Sugar 196.86 g → 160.30g

Lemon Juice 193.37 g → 183.69g

The lemon juice will be the first ingredient to be used up, there will be a surplus of water and sugar.

I hope this helps!

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2 years ago

How much maleic anhydride would you need to react 178 mg of anthracene? Assume 1:1 ratio from maleic anhydride to anthracene.

Licemer1 [7]

Answer:

(1) 0.10 (2) 17.8 g

Explanation:

Since the reaction ratio is 1:1 what we need is to convert the given masses to moles and you will have the answer:

MW anthracene = 178.23 g/mol

MW maleic anhydride = 98.06 g/mol

a) mass anthracene = 178 mg x 1 g/ 1000 mg = 0.178 g anthracene

Moles anthracene = 0.178 g anthracene/ 178.23 g/mol

= 0.001 mol anthracene

0.001 mol anthracene x 1 mol maleic acid/mol anthracene

= 0.001 mol maleic anhydride

mass maleic anhydride = 0.001 mol x 98.06 g/mol = 0.10 g

b) moles maleic anhydride = 9.8 g/ 98.06 g/mol = 0.099 moles

0.099 moles maleic anhydride x 1 mol anthracene/mol maleic anhydride =

0.099 mol anthracene

g anthracene = 0.10mol x 178 g/mol = 17.8 g

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2 years ago

What is the hydronium ion concentration of a solution with a pOH of 7.20?

-Dominant- [34]

Answer:

[H⁺] = 1.58 x 10⁻⁷ M.

Explanation:

∵ pOH = - log[OH⁻]

7.20 = - log[OH⁻]

log[OH⁻] = - 7.20

∴ [OH⁻] = 6.31 x 10⁻⁸.

∵ [H⁺][OH⁻] = 10⁻¹⁴.

∴ [H⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(6.31 x 10⁻⁸) = 1.585 x 10⁻⁷ M.

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2 years ago

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A student pours exactly 26.9 mL of HCl acid of unknown molarity into a beaker. The student then adds 2 drops of the indicator an

Assoli18 [71]

a.
Acids react with bases and give salt and water and the products.

Hence, HCl reacts with NaOH and gives NaCl salt and H₂O as the products. The reaction is,
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

To balance the reaction equation, both sides hould have same number of elements.

Left hand side, Right hand side,
H atoms = 2 H atoms = 2
Cl atoms = 1 Cl atoms = 1
Na atoms = 1 Na atoms = 1
O atoms = 1 O atoms = 1

Hence, the reaction equation is already balanced.

b.
Molarity (M)= moles of solute (mol) / Volume of the solution (L)

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Molarity of NaOH = 0.13 M
Volume of NaOH added = 43.7 mL
Hence, moles of NaOH added = 0.13 M x 43.7 x 10⁻³ L
= 5.681 x 10⁻³ mol

Stoichiometric ratio between NaOH and HCl is 1 : 1

Hence, moles of HCl = moles of NaOH
= 5.681 x 10⁻³ mol

5.681 x 10⁻³ mol of HCl was in 26.9 mL.

Hence, molarity of HCl = 5.681 x 10⁻³ mol / 26.9 x 10⁻³ L
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2 years ago

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Bad White [126]

Answer:

green traveled les distance but still ended up in the same location as red

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2 years ago

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