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azamat
3 years ago
12

Which method is used to separate parts of a solution when the entire solution can pass through a filter?

Chemistry
2 answers:
SCORPION-xisa [38]3 years ago
6 0

Answer:Distillation screening method

Explanation:

wel3 years ago
3 0
Distillation screening method
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Analysis of the water content of a lake found in the desert showed that it contained 17.5 percent chloride ion, and has a densit
tankabanditka [31]

Answer:

The molarity is 6.0 M.

Explanation:

<em>The </em><em>concentration</em><em> of a solution is the amount of solute present in a given amount  of solvent, or a given amount of solution.</em>

In this problem, we are asked to convert percent by mass to molarity, which are two different concentration units.

<em>The </em><em>percent by mass </em><em>is the ratio of the mass of a solute to the mass of the solution, multiplied by 100 percent.</em> On the other hand, <em>molarity (M)</em><em> is defined as the number of moles of solute per liter of solution.</em>

If we have 17.15 percent chloride ion, that means that we have 17.15 grams of chloride ion in 100 grams of solution. The density provided is the density of the solution, so we calculate how many mL correspond to 100 grams of solution:

1.23 g ----------- 1 mL

100 g ------------ <u>x= 81.3 mL</u>

Therefore, 17.5 grams of chloride ion are contained in 81.3 mL. Now we need to convert these grams into moles.

The atomic mass of chlorine is 35.5 g/mol:

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

17.5 g --------------- <u>x= 0. 49 mol</u>

This 0.49 moles are contained in 81.13 mL, the definition of molarity says that this moles are contained in a liter (or what is the same, 1000 mL):

81.3 mL --------------- 0.49 mol

1000 mL ------------- x= 6.0 M

The molarity of the chloride ion will be 6.0 M.

5 0
3 years ago
A 50.0 mL sample of an aqueous H2SO4 solution is titrated with a 0.375 M NaOH solution. The equivalence point is reached with 62
vesna_86 [32]

Answer: The concentration of H_2SO_4 is 0.234 M

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity H_2SO_4 = 2

M_1 = molarity of H_2SO_4 solution = ?

V_1 = volume of  H_2SO_4 solution = 50.0 ml

n_2 = acidity of NaOH = 1

M_1 = molarity of NaOH solution = 0.375 M

V_1 = volume of  NaOH solution = 62.5 ml

Putting in the values we get:

2\times M_1\times 50.0=1\times 0.375\times 62.5

M_1=0.234M

Therefore concentration of H_2SO_4 is 0.234 M

6 0
3 years ago
How many molecules of XeF6 are formed from 12.9 L of F2 (at 298 K and 2.6 atm) according to 11) the following reaction? Assume t
ddd [48]

Answer:

#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.

Explanation:

Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆

1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.

=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP

2. Calculate moles of F₂ used

=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used

3. Calculate moles of XeF₆ produced from reaction ratios …

Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆

4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number  (6.02 x 10²³ molecules/mole)

=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)

                                  = 2.75 x 10²³ molecules XeF₆.

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Density= Mass/Volume I am positive I just had an assignment on this
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