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Ronch [10]
3 years ago
14

The molecule with only single covalent bonds is _. (C) * CO2 СО CI2 ON2

Chemistry
2 answers:
bezimeni [28]3 years ago
5 0
(C12) step by step answers
anyanavicka [17]3 years ago
3 0

Answer:

CO (carbon monoxide)

Explanation:

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I am so lost, and suggestions to how I can solve this?
Ad libitum [116K]
<h3>Answer:</h3>

165.078 g

<h3>Explanation:</h3>

<u>We are given;</u>

  • Specific heat capacity of nickel as 0.446 J/g°C
  • Initial temperature of nickel is 255.5 °C
  • Mass of water in the calorimeter is 250.0 g
  • Initial temperature of water is 20° C
  • Final temperature of the mixture is 35.5°C
  • Specific heat capacity of water is 4.18 J/g°C

We are required to calculate the mass of nickel sample;

<h3>Step 1: Calculate the amount of heat absorbed by water.</h3>

We know that quantity of heat absorbed, Q = m × c × ΔT

Change in temperature, ΔT = 35.5°C - 20°C

                                              = 15.5 °C

Therefore, Q = 250 g × 4.1 J/g°C × 15.5° C

                     = 16,197.5 Joules

<h3>Step 2: Calculate the amount of heat released by nickel sample</h3>

Assuming the mass of nickel sample is m

Then, Heat released, Q = m × c × ΔT

Change in temperature, ΔT = 255.5 °C - 35.5 ° C

                                              = 220 °C

Q = m × 0.446 J/g°C × 220° C

   = 98.12m joules

<h3>Step 3: Determine the mass of nickel sample</h3>

We know that the amount of heat absorbed is equivalent to amount of heat released.

That is, Quantity of heat absorbed = Quantity of heat released

Therefore;

98.12 m joules = 16,197.5 Joules

    m = 165.078 g

Thus, the mass of nickel sample is 165.078 g

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4 years ago
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Light bending as it passes through a rain drop is an example of Transmission Diffraction Refraction Reflection
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3 years ago
Calculate the solubility in g/l of lead (ii) fluoride in water at 25 oc if ksp of pbf2 is 4. 1 x 10-8.
Svetlanka [38]

The solubility of Lead(II)Fluoride is 2.17 × 10⁻³ g/L in water at 25°C.

At a specific solution temperature, a solid salt compound can entirely dissolve in pure water up to a predetermined molar solubility limit. The dissociation stoichiometry ensures that the molarities of the constituent ions are proportionate to one another. The saturable nature of the solution causes them to also coexist in a solubility equilibrium with the solid component. At this temperature, a solubility product constant Ksp is calculated using the solubility product of their molarity values.

Lead (II) fluoride has the following solubility equilibrium for its saturated solution:

                   PbF_2(s) ⇄ Pb^2^+ (aq) + 2F^- (aq)

                   K_s_p = [Pb^2^+][F^-]^2

This compound dissociates in a 1:2 ratio of ions. For the compound dissolved in pure water, the Ksp is expressed in terms of the molar solubility "x" as:

K_s_p = (x) (2x)^2\\

K_s_p = 4x^3

Here,

K_s_p= 4.1 × 10^-^8

4.1 × 10⁻⁸ = 4 x³

x³ = 1.025 × 10⁻⁸

x³ = 10.25 × 10⁻⁹

x = 2.17 × 10⁻³ g/L

Therefore, the solubility of Lead(II)Fluoride is 2.17 × 10⁻³ g/L.

Learn more about solubility here:

brainly.com/question/23946616

#SPJ4

6 0
2 years ago
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