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postnew [5]
3 years ago
15

If 75.0 g of a liquid has a volume of 62.4 mL ,calculate the liquid’s density

Chemistry
1 answer:
Galina-37 [17]3 years ago
4 0
<span>To calculate the density of a liquid, you have to first know that density is the amount of substance per unit of volume. In this specific question, density will be found with units of g/mL. Now, the density can be found by dividing the amount of liquid, 75.0g, by the volume, 62.4mL. Doing this we get: 75.0g/62.4mL= 1.2 g/mL as the density of the liquid.</span>
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Pls helps asap plsssssss
Lisa [10]

Answer:

A spectator ion refers to a charged atom in a chemical reaction that does not undergo a chemical change.

Hope this helps!!!

4 0
2 years ago
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A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would
Olenka [21]

Answer:

C. Ca_3(PO_4)_2  will precipitate out first

the percentage of Ca^{2+}remaining =  12.86%

Explanation:

Given that:

A solution contains:

[Ca^{2+}] = 0.0440 \ M

[Ag^+] = 0.0940 \ M

From the list of options , Let find the dissociation of Ag_3PO_4

Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}

where;

Solubility product constant Ksp of Ag_3PO_4 is 8.89 \times 10^{-17}

Thus;

Ksp = [Ag^+]^3[PO_4^{3-}]

replacing the known values in order to determine the unknown ; we have :

8.89 \times 10 ^{-17}  = (0.0940)^3[PO_4^{3-}]

\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}  = [PO_4^{3-}]

[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}

[PO_4^{3-}] =1.07 \times 10^{-13}

The dissociation  of Ca_3(PO_4)_2

The solubility product constant of Ca_3(PO_4)_2  is 2.07 \times 10^{-32}

The dissociation of Ca_3(PO_4)_2   is :

Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}

Thus;

Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2

2.07 \times 10^{-33} = (0.0440)^3  [PO_4^{3-}]^2

\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}=   [PO_4^{3-}]^2

[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}

[PO_4^{3-}]^2 = 2.43 \times 10^{-29}

[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}

[PO_4^{3-}] =4.93 \times 10^{-15}

Thus; the phosphate anion needed for precipitation is smaller i.e 4.93 \times 10^{-15} in Ca_3(PO_4)_2 than  in  Ag_3PO_4  1.07 \times 10^{-13}

Therefore:

Ca_3(PO_4)_2  will precipitate out first

To determine the concentration of [Ca^+] when  the second cation starts to precipitate ; we have :

Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2

2.07 \times 10^{-33}  = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2

[Ca^{2+}]^3 =  \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}

[Ca^{2+}]^3 =1.808 \times 10^{-7}

[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}

[Ca^{2+}] =0.00566

This implies that when the second  cation starts to precipitate ; the  concentration of [Ca^{2+}] in the solution is  0.00566

Therefore;

the percentage of Ca^{2+}  remaining = concentration remaining/initial concentration × 100%

the percentage of Ca^{2+} remaining = 0.00566/0.0440  × 100%

the percentage of Ca^{2+} remaining = 0.1286 × 100%

the percentage of Ca^{2+}remaining =  12.86%

5 0
3 years ago
point Deshawn ran the 400 m race(in a straight line) in 2 minutes. What is his distance and displacement ? Explain your answer b
san4es73 [151]

His distance and displacement are the same, which was 400 m

<h3>Further explanation</h3>

Given

Distance = 400 m

time = 2 min

Required

Distance and displacement

Solution

Distance is a scalar quantity that indicates the length of the trajectory that is traveled by an object within a certain interval. Distance has no direction, only has magnitude  

Can be simplified distance = totals traveled  

Displacement is a vector quantity that shows changes in the position of objects in a certain interval of time. Displacement has magnitude and direction  

Can be simplified displacement = distanced traveled from starting point to ending point  

From the definition above shows that the displacement and the distance that he traveled have the same value (magnitude), which is equal to 400 m

The value of the two will be different if he starts and finishes at the same point, then the displacement value is zero while the distance he has traveled is still 0

4 0
3 years ago
When a drawing of a circle made in the center of a piece of paper with a black marker gets wet, the marker bleeds, and as the wa
suter [353]

Answer:

D) Chromatography

Explanation:

8 0
3 years ago
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The sp of pbbr2 is 6. 60×10−6. what is the molar solubility of pbbr2 in pure water?
Inessa05 [86]

Ksp of PbBr₂ is 6.60 × 10⁻⁶. The molar solubility of PbBr₂ in pure water is 0.0118M.

Ksp or Solubility Product Constant is an equilibrium constant for the dissociation in an aqueous solution.

Molar solubility (S) is the concentration of the dissolved substance in a solution that is saturated.

Let the molar solubility be S upon dissociation.

PbBr₂ or Lead Bromide dissociates in pure water as follows:

                          PbBr₂ ----------> Pb⁺² + Br⁻

                                                     S      2S

Ksp = [Pb⁺²] [ Br⁻]

Ksp = (S) (2S)²

Ksp = 4S³

6.60 × 10⁻⁶ = 4S³

S = 0.0118M

Hence, the Molar solubility S is 0.0118M.

Learn more about Molar solubility here, brainly.com/question/16243859

#SPJ4

6 0
2 years ago
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