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meriva
3 years ago
9

Ethyl alcohol (ch3ch2oh) is/is not soluble in water. 1. is; all organic molecules are soluble in water. 2. is; ethyl alcohol exh

ibits dipole-dipole and h-bonding interactions with water. 3. is not; ethyl alcohol exhibits dispersion forces which are insufficient for dissolving. 4. is not; ethyl alcohol is an ionic compound with a hydroxide group and hydroxides are insoluble.
Chemistry
1 answer:
nata0808 [166]3 years ago
5 0
Answer:
            Ethyl alcohol is soluble in water because <span>ethyl alcohol exhibits dipole-dipole and h-bonding interactions with water.

Explanation:
                   Ethyl alcohol and water are miscible in each other because both are polar in nature and "Like dissolves Like".
                   The bond between oxygen and hydrogen atoms, both in alcohol and water are polar in nature and results in intermolecular hydrogen bond interactions between them as hydrogen bonding results when hydrogen atom in one molecule directly attached to highly electronegative atoms like fluorine, oxygen and nitrogen forms interaction with higly electronegative atom of neighbor atom.</span>
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6 0
3 years ago
The ksp of copper(ii) carbonate, cuco3, is 1.4 × 10-10. calculate the molar solubility of this compound.
mafiozo [28]
As,
                              CuCO₃    ⇆  Cu²⁺  +  CO₃²⁻
So,
                             Kc  =  [Cu²⁺] [CO₃²⁻] / CuCO₃
Or,
                             Kc (CuCO₃)  =  [Cu²⁺] [CO₃²⁻]
Or,
                             Ksp  =  [Cu²⁺] [CO₃²⁻]
As,
Ksp  = 1.4 × 10⁻¹⁰
So,
                            1.4 × 10⁻¹⁰  =  [x] [x]
Or,
                             x²  =  1.4 × 10⁻¹⁰ 
Or,
                             x  =  1.18 × 10⁻⁵ mol/L
To cahnge ito g/L,
                             x  =  1.18 × 10⁻⁵ mol/L × 123.526 g/mol

                   
         x  =  1.45 × 10⁻³ g/L
3 0
3 years ago
Calculate the [OH-] given pH of 9.9
Juli2301 [7.4K]

Answer

pOH = 4.1

Explanation

<em>Given:</em>

pH = 9.9

<em>Required</em>: The concentration of OH-

Solution

pH + pOH = 14

9.9 + pOH = 14

pOH = 14-9.9

pOH = 4.1

3 0
1 year ago
Wendy is using a poorly calibrated electronic balance to measure the mass of a crucible. Her technique in making the measurement
Nataly [62]
B answer is b I repeat it’s b
4 0
3 years ago
(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak
Taya2010 [7]

Answer:

Here's what I get  

Explanation:

Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We must calculate the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's gather all the information in one place.

                   H₂ +    I₂    ⇌ 2HI

I/mol·L⁻¹:    0.30   0.15         x

2. Calculate the concentration of HI

Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} =  5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}

3. Plot the initial points

The graph below shows the initial concentrations plotted on the vertical axis.

 

7 0
3 years ago
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