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pav-90 [236]
3 years ago
15

What is the density of a object whose dimensions are 5.54

Chemistry
1 answer:
Tcecarenko [31]3 years ago
5 0

The object with dimensions 5.54\ cm \times 10.6\ cm \times 199\ cm and with mass 250.6 g will have a density equal to 0.02144\ g/cm^3.

<h3><u>Explanation:</u></h3>

<u>Given:</u>

Dimensions of the object = 5.54\ cm \times 10.6\ cm \times 199\ cm

length = l = 5.54 cm

width = w = 10.6 cm

height = h = 199 cm

Mass of the object = m = 250.6 g

<u>Formula:</u>

The relation between density, mass and volume of any object is given by:

Density = \frac{ Mass }{Volume}

Since volume of an object with length (l), width (w) and height (h) [assuming a rectangular object] can be defined as,

V = l \times w \times h =5.54 \times 10.6 \times 199

Volume = V = 11686.076\ cm^3

Density = Mass / Volume = \frac{250}{11686.076}

Density = 0.02144 g/cm^3

Density of the object is 0.02144 g/cm^3

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Colt1911 [192]

Answer:

Moles of silver iodide produced = 1.4 mol

Explanation:

Given data:

Mass of calcium iodide = 205 g

Moles of silver iodide produced = ?

Solution:

Chemical equation:

CaI₂ + 2AgNO₃     →      2AgI + Ca(NO₃)₂

Number of moles calcium iodide:

Number of moles = mass/ molar mass

Number of moles = 205 g/ 293.887 g/mol

Number of moles = 0.7 mol

Now we will compare the moles of calcium iodide with silver iodide.

                     CaI₂         :           AgI

                         1           :             2

                       0.7         :           2×0.7 = 1.4

Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.

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3 years ago
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Consider the following reaction at equilibrium. 2CO2 (g) 2CO (g) + O2 (g) H° = -514 kJ Le Châtelier's principle predicts that th
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Answer:

C. at low temperature and low pressure.

Explanation:

  • <em>Le Châtelier's principle </em><em>states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>

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  • For the reaction:

<em>2CO₂(g) ⇄ 2CO(g) + O₂(g), ΔH = -514 kJ.</em>

<em></em>

<em><u>Effect of pressure:</u></em>

  • When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
  • The reactants side (left) has 2.0 moles of gases and the products side (right) has 3.0 moles of gases.

<em>So, decreasing the pressure will shift the reaction to the side with higher no. of moles of gas (right side, products), </em><em>so the equilibrium partial pressure of CO (g) can be maximized at low pressure.</em>

<em></em>

<u><em>Effect of temperature:</em></u>

  • The reaction is exothermic because the sign of ΔH is (negative).
  • So, we can write the reaction as:

<em>2CO₂(g) ⇄ 2CO(g) + O₂(g) + heat.</em>

  • Decreasing the temperature will decrease the concentration of the products side, so the reaction will be shifted to the right side to suppress the decrease in the temperature, <em>so the equilibrium partial pressure of CO (g) can be maximized at low temperature.</em>

<em></em>

  • So, the right choice is:

<em>C. at low temperature and low pressure.</em>

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7 0
4 years ago
The boiling point elevation of an aqueous sucrose solution is found to be 0.39°C.
SVEN [57.7K]

Answer:

130.4 grams of sucrose, would be needed to dissolve in 500 g of water.

Explanation:

Colligative property of boiling point elevation:

ΔT = Kb . m . i

In this case, i = 1 (sucrose is non electrolytic)

ΔT = Kb . m

0.39°C = 0.512°C/m . m

0.39°C /0.512 m/°C = m

0.762 m (molality means that this moles, are in 1kg of solvent)

If in 1kg of solvent, we have 0.712 moles of sucrose, in 500 g, which is the half, we should have, the hallf of moles, 0.381 moles

Molar mass sucrose = 342.30 g/m

Molar mass . moles = mass

342.30 g/m . 0.381 m = 130.4 g

6 0
4 years ago
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