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dimaraw [331]
4 years ago
10

If 287 mL of a 1.6 M glucose solution is diluted to 412 mL , what is the molarity of the diluted solution

Chemistry
1 answer:
satela [25.4K]4 years ago
5 0

Answer:

The answer to your question is 1.11 M

Explanation:

Data

volume 1 = 287 ml

concentration 1 = 1.6 M

volume 2= 412 ml

concentration 2 = ?

Formula

               Volume 1 x concentration 1 = Volume 2 x concentration 2

Solve for concentration 2

          concentration 2 = (volume 1 x concentration 1) / volume 2

Substitution

          concentration 2 = (287 x 1.6) / 412

Simplification

          concentration 2 = 459.2 / 412

Result

           concentration 2 = 1.11 M  

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0.65743 In scientific notation
Goryan [66]
202827.0000 is the answer I think idrk
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3 years ago
What is the coefficient of mno2 when you balance the equation for this redox reaction? kcn(aq) + kmno4(aq) + h2o(l) → kcno(aq) +
Tasya [4]
Kcn(aq)+kmno4(aq)+h2o(l)----->2mno2(s)+koh(aq)

7 0
3 years ago
Read 2 more answers
Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?
s2008m [1.1K]

Answer:

65.08 g.

Explanation:

  • For the reaction, the balanced equation is:

<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

<u><em>Using cross multiplication:</em></u>

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol ) = <em>65.08 g.</em>

4 0
3 years ago
Can you please help me and can you show your work please
natali 33 [55]

Answer:

1. 25 moles water.

2. 41.2 grams of sodium hydroxide.

3. 0.25 grams of sugar.

4. 340.6 grams of ammonia.

5. 4.5x10²³ molecules of sulfur dioxide.

Explanation:

Hello!

In this case, since the mole-mass-particles relationships are studied by considering the Avogadro's number for the formula units and the molar mass for the mass of one mole of substance, we proceed as shown below:

1. Here, we use the Avogadro's number to obtain the moles in the given molecules of water:

1.5x10^{25}molecules*\frac{1mol}{6.022x10^{23}molecules} =25 molH_2O

2. Here, since the molar mass of NaOH is 40.00 g/mol, we obtain:

1.2mol*\frac{40.00g}{1mol} =41.2g

3. Here, since the molar mass of C6H12O6 is 180.15 g/mol:

45g*\frac{1mol}{180.15g}=0.25g

4. Here, since the molar mass of ammonia is 17.03 g/mol:

20mol*\frac{17.03g}{1mol}=340.6g

5. Here, since the molar mass of SO2 is 64.06 g/mol:

48g*\frac{1mol}{64.06g} *\frac{6.022x10^{23}molecules}{1mol} =4.5x10^{23}molecules

Best regards!

5 0
3 years ago
Filtration, or screening, is one way to separate certain mixtures, as shown in the image. Which of these mixtures could be most
erastovalidia [21]

Answer: D

Explanation:

3 0
3 years ago
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