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Ivenika [448]
3 years ago
12

What criteria must be met for a collision to result in particles changing?

Chemistry
1 answer:
musickatia [10]3 years ago
7 0
Sufficient energy upon collision of particles and correct orientation of particles
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A student carefully pipets 10.00ml of ethanol into a beaker. If ethanol has a density of 0.779g/ml, how many grams of ethanol we
lesantik [10]

Answer: 7.79 grams of ethanol were put into the beaker.

Explanation:

To calculate the mass of ethanol, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of ethanol = 0.779 g/mL

Volume of water = 10.00 mL

Putting values in above equation, we get:

0.779g/mL=\frac{\text{Mass of ethanol}}{10.00mL}\\\\\text{Mass of ethanol}=(0.779g/mL\times 10.00mL)=7.79g

Thus 7.79 grams of ethanol were put into the beaker.

5 0
3 years ago
An analytical chemist is titrating 118.3 mL of a 0.3500 M solution of butanoic acid (HC3H7CO2) with a 0.400 M solution of KOH. T
TEA [102]

Answer:

pH = 12.33

Explanation:

Lets call HA = butanoic acid and A⁻ butanoic acid and its conjugate base butanoate respectively.

The titration reaction is

HA + KOH ---------------------------- A⁻ + H₂O + K⁺

number of moles of HA :   118.3 ml/1000ml/L x 0.3500 mol/L = 0.041 mol HA

number of  moles of OH  : 115.4 mL/1000ml/L x 0.400 mol/L  = 0.046 mol A⁻

therefore the weak acid will be completely consumed and what we have is  the unreacted strong base KOH which will drive the pH of the solution since the contribution of the conjugate base is negligible.

n unreacted KOH = 0.046 - 0.041 = 0.005 mol KOH

pOH = - log (KOH)

M KOH = 0.005 mol / (0.118.3 +0.1154)L = 0.0021 M

pOH = - log (0.0021) = 1.66

pH = 14 - 1.96 = 12.33

Note: It is a mistake to ask for the pH of the <u>acid solutio</u>n since as the above calculation shows we have a basic solution the moment all the acid has been consumed.

4 0
2 years ago
Which statement best describes the domain and range of f(x) = -(7)X and g(x) = 7X7 O f(x) and g(x) have the same domain and the
Ivan

Answer:

The graph of this equation is shown in Figure 1. As you can see this is a straight line with negative slope and does not intersect the y-axis. So the ...

Explanation:

4 0
2 years ago
The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the s
raketka [301]

Answer : The equilibrium concentration of H_3O^+ in the solution is, 2.1\times 10^{-3}M

Explanation :

The dissociation of acid reaction is:

                       C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-

Initial conc.        c                                 0                0

At eqm.             c-x                                 x                x

Given:

c = 7.0\times 10^{-2}M

K_a=6.3\times 10^{-5}

The expression of dissociation constant of acid is:

K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}

K_a=\frac{(x)\times (x)}{(c-x)}

Now put all the given values in this expression, we get:

6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}

x=2.1\times 10^{-3}M

Thus, the equilibrium concentration of H_3O^+ in the solution is, 2.1\times 10^{-3}M

4 0
3 years ago
Equal molar quantities of Ca2 and EDTA (H4Y) are added to make a 0.010 M solution of CaY2- at pH 10. The formation constant for
abruzzese [7]

Answer:

the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

Explanation:

Given the data in the question;

Ca^{2+ + y^{4- ⇄  CaY^{2-

Formation constant Kf

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) = 5.0 × 10¹⁰

Now,

[y^{4-] = \alpha _4CH_4Y; ∝₄ = 0.35

so the equilibrium is;

Ca^{2+ + H_4Y ⇄  CaY^{2- + 4H⁺

Given that; CH_4Y = Ca^{2+     { 1 mol Ca^{2+  reacts with 1 mol H_4Y  }

so at equilibrium, CH_4Y = Ca^{2+ = x

∴

Ca^{2+ + y^{4- ⇄  CaY^{2-

x        + x         0.010-x

since Kf is high, them x will be small so, 0.010-x is approximately 0.010

so;

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) =  CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ 0.010 / ( [x][ 0.35 × x] )  = 5.0 × 10¹⁰

⇒ 0.010 / 0.35x²  = 5.0 × 10¹⁰

⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 5.7142857 × 10⁻¹³

⇒ x = √(5.7142857 × 10⁻¹³)

⇒ x = 7.559 × 10⁻⁷

Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

8 0
2 years ago
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