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Anni [7]
3 years ago
6

The density of lead is 11.3g/cm^3 and the density of chromium is 7.14g/cm^3. If we have one gram of each metal, what is the rati

o of the volume of lead to the chromium?
Chemistry
1 answer:
Lady bird [3.3K]3 years ago
5 0

The ratio of the volume of lead to the volume of chromium is 22 : 35

Density is simply defined as the mass of a substance per unit volume of the substance.

<h3>Density = mass / volume </h3>

To solve the question given above, we'll begin by calculating the volume of lead and chromium. This can be obtained as follow:

<h3>For Lead:</h3>

Mass of lead = 1 g

Density of lead = 11.3 g/cm³

<h3>Volume of lead =? </h3>

Density = mass /volume

11.3 = 1 / Volume

Cross multiply

11.3 × Volume = 1

Divide both side by 11.3

Volume = 1 / 11.3

<h3>Volume of lead = 0.088 cm³</h3>

<h3>For Chromium:</h3>

Mass of chromium = 1 g

Density of chromium = 7.14 g/cm³

<h3>Volume of chromium =? </h3>

Density = mass /volume

7.14 = 1 / Volume

Cross multiply

7.14 × Volume = 1

Divide both side by 7.14

Volume = 1 / 7.14

<h3>Volume of chromium = 0.14 cm³</h3>

Finally, we shall determine the ratio of the volume of lead to the volume chromium. This can be obtained as follow:

Volume of lead = 0.088 cm³

Volume of chromium = 0.14 cm³

<h3>Ratio of the volume of lead to chromium =? </h3>

Ratio = Volume of lead / Volume of chromium

Ratio = 0.088 / 0.14

Ratio = 22 / 35

<h3>Ratio = 22 : 35</h3>

Therefore, the ratio of the volume of lead to the volume chromium is 22 : 35

Learn more: brainly.com/question/17262276

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6 0
2 years ago
The equation shows one mole of ethanol fuel being burned in oxygen. Convert the energy released into its equivalent mass. C2H5OH
lana [24]

Answer:

47.9 g of ethanol

Explanation:

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Since 1 mole of ethanol when combusted releases 1367 kJ/mol of energy

x moles of ethanol releases 1418 kJ/mol.

x= 1 × 1418 kJ/mol/ 1367 kJ/mol

x= 1.04 moles of ethanol.

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In a hot shower or steam room, you look up at the ceiling. What do you observe on the ceiling and why?
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3 years ago
A saturated solution of manganese(II) hydroxide was prepared, and an acid–base titration was performed to determine its KspKsp a
trapecia [35]

Answer:

10.945 x 10^-4

Explanation:

Balanced equation:

Mn(OH)2 +  2 HCl  -->  MnCl2  +  H2O

it takes 2 moles HCL for each mole Mn(OH)2

Next find the molarity of the Mn(OH)2 solution

= (1 mole Mn(OH)2 / 2 mole HCl)  X (0.0020 mole HCl / 1000ml) X (4.86 ml)    

= 4.86 x 10^-3 mole  

this is now dissolved in (70 + 4.86)  =  74.86 ml or 0.07486 L

thus [Mn(OH)2]  =  4.86 x 10^-3 mole / 0.07486 L  =  0.064921 M

Ksp =  [Mn2+][OH-]^2  =  4x^3  =  4(0.064921)^3  = 10.945 x 10^-4

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