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LuckyWell [14K]
2 years ago
7

Dominic made the table below to organize his notes about mixtures. A 1-column table. The first column labeled properties of mixt

ures has entries has no set composition, must have more than one state of matter, must have more than one substance. What mistake did Dominic make? The title should read “Properties of Solutions” because some mixtures do not have all of the properties listed. There is a definite recipe to make each mixture, so the composition of a mixture is set. Although it is possible to have more than one state, it is also possible to have only one state. A single substance can be used to make a mixture if the substance is composed of more than one element.
Chemistry
2 answers:
Vinvika [58]2 years ago
5 0

Answer:

hmmm

Explanation:

lettt meee thinkkk abouttt itt

vodomira [7]2 years ago
3 0

Answer:

Although it is possible to have more than one state, it is also possible to have only one state.

Explanation:

Took the unit test

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Solutions can be composed of:
Anika [276]

A solution may exist in any phase so your answer is D. any of the above

hope this helps :)

7 0
3 years ago
During a synthesis reaction, 1.8 grams of magnesium reacted with 6.0 grams of oxygen. What is the maximum amount of magnesium ox
skelet666 [1.2K]

Answer:

2.9 grams.

Explanation:

  • From the balanced reaction:

<em>Mg + 1/2O₂ → MgO,</em>

1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.

  • We need to calculate the no. of moles of (1.8 g) of Mg and (6.0 g) of oxygen:

no. of moles of Mg = mass/molar mass = (1.8 g)/(24.3 g/mol) = 0.074 mol.

no. of moles of O₂ = mass/molar mass = (6.0 g)/(16.0 g/mol) = 0.375 mol.

<em>So. 0.074 mol of Mg reacts completely with (0.074/2 = 0.037 mol) of O₂ which be in excess.</em>

<em></em>

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

1.0 mole of Mg produce → 1.0 mol of MgO.

∴ 0.074 mol of Mg produce → 0.074 mol of MgO.

<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.074 mol)(40.3 g/mol) = <em>2.98 g.</em>

5 0
3 years ago
Two samples of sodium chloride with different masses were decomposed into their constituent elements. One sample produced 1.731.
Amanda [17]

Answer:

The answer to your question is letter c) 6.09 g of sodium and 9.38 g of chlorine.

Explanation:

This problem is solve using rule of three

We know that the proportion Sodium to Chloride is 1 to 1 in sodium chloride, so we have to look for this proportion in the options

AM Sodium = 23 g

AM Chlorine = 35.5 g

                 Sodium                                     Chlorine

           23 g ---------------- 1 mol              35.5 g -------------- 1 mol

       1713.73g -------------    x               2666.6 g -------------    x

          x = 1713.73/23 = 74.51                     x = 2666.6/35.5 = 75.12

    These values are very similar, we have to look for the proportion in the options

a)      6.09g of sodium = 0.26 mol

      4,87 g of chlorine = 0.14 mol           These numbers are not very similar

b) We have 0.26 mol of Na

                  0.037 mol of Cl                      This is not the answer

c) We have 0.26 mol of Na

                  0.26 mol of Cl                     These numbers are the same, the proportion is 1:1, this is the answer

d) We have 0.26 mol of Na

                   0.36 mol of Cl                    This is not the answer

8 0
3 years ago
In the titration of 50. 0 mL of 0. 400 M HCOOH with 0. 150 M LiOH, how many mL of LiOH are required to reach the equivalence poi
mart [117]

The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL

<h3>Balanced equation </h3>

HCOOH + LiOH —> HCOOLi + H₂O

From the balanced equation above,

The mole ratio of the acid, HCOOH (nA) = 1

The mole ratio of the base, LiOH (nB) = 1

<h3>How to determine the volume of LiOH </h3>
  • Molarity of acid, HCOOH (Ma) = 0.4 M
  • Volume of acid, HCOOH (Va) = 50 mL
  • Molarity of base, LiOH (Mb) = 0.15 M
  • Volume of base, LiOH (Vb) =?

MaVa / MbVb = nA / nB

(0.4 × 50) / (0.15 × Vb) = 1

20 / (0.15 × Vb) = 1

Cross multiply

0.15 × Vb = 20

Divide both side by 0.15

Vb = 20 / 0.15

Vb = 133.3 mL

Thus, the volume of the LiOH solution needed is 133.3 mL

Learn more about titration:

brainly.com/question/14356286

8 0
1 year ago
Which of the following elements has the largest ionization
Tanya [424]

Li

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8 0
3 years ago
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