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Nookie1986 [14]
3 years ago
5

which of the following chemical equilibrium equations best shows what happens in the buffer solutions to minimize the change in

ph when a small amount of a strong base is added?
Chemistry
1 answer:
Viefleur [7K]3 years ago
4 0

Answer:

HCO3−(aq)+ OH−(aq) ⇄ CO32−(aq) + H2O(l)

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Assignment<br>Explain why it is not advisable to wear<br>black silky cloth in the sun​
Crazy boy [7]

Answer:

It is not water absorbant and also the colour black, due to it's dark nature makes the person wearing the cloth feel hot. Unlike cotton, silk is not at all water absorbant as it does not absord any sweat produced in the wearers body

3 0
3 years ago
If 0.896 g of a gas occupies a 250 mL flask at 20°C and 760 mm Hg of pressure, what is the molar mass of the gas?​
salantis [7]

Answer:

86.2 g/mol

Explanation:

Before you can find the molar mass, you first need to calculate the number of moles of the gas. To find this value, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (mmHg)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (62.36 L*mmHg/mol*K)

-----> T = temperature (K)

After you convert the volume from mL to L and the temperature from Celsius to Kelvin, you can use the equation to find the moles.

P = 760 mmHg                                      R = 62.36 L*mmHg/mol*K

V = 250 mL / 1,000 = 0.250 L              T = 20 °C + 273.15 = 293.15 K

n = ? moles

PV = nRT

(760 mmHg)(0.250 L) = n(62.36 L*mmHg/mol*K)(293.15 K)

190 = n(18280.834)

0.0104 = n

The molar mass represents the mass (g) of the gas per every 1 mole. Since you have been given a mass and mole value, you can set up a proportion to determine the molar mass.

\frac{?grams}{1 mole} =\frac{0.896grams}{0.0104moles}                                      <----- Proportion

?grams(0.0104moles) = 0.896                       <----- Cross-multiply

?grams = 86.2                                               <----- Divide both sides by 0.0104

8 0
2 years ago
An analysis reveals that a substance consists of 26.7% of carbon, 2.24% of hydrogen and 71.1% of oxygen, If the molar mass of th
exis [7]
 the   molecular  formula    is  calculated  as  follows 
find  the  mole  of  each  molecule 
that  is   for  carbon = 26.7/12=  2.23  moles
                   hydrogen=  2.24/1=2.24  moles
                 oxygen=  71.1/16=4.44  moles
find  the   mole  ratio  of  each  element
that  is  divide  all  moles  by  the  smallest  mole(  2.23)

for  carbon=2.23/2.23=1,
  hydrogen =  2.24/2,23=1
  oxygen= 4.44/2.23=  2
the   empirical  formula  = CHO2
therefore  the  molecular  formula=(CHO2)n=270.1
                 { ( 12x 1)  + 1x1) +(16x2)}n=270.1
45n=270.1 
n=6
molecular  formula  is  therefore=(CHO2)6=C6H6O12

7 0
3 years ago
An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +
Maurinko [17]

Answer:

The simplified expression for the fraction  is  \text {X} =    \dfrac{  {k_3  \times cM} }{k_1 +k_2 + k_3 }

Explanation:

From the given information:

O3* → O3                   (1)    fluorescence

O + O2                      (2)    decomposition

O3* + M → O3 + M    (3)     deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

\text {X} =    \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) }  } }

\text {X} =    \dfrac{  {k_3 \times cO \times cM} }{  {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) }  }

\text {X} =    \dfrac{  {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3  \times cM) }

\text {X} =    \dfrac{  {k_3  \times cM} }{k_1 +k_2 + k_3  }    since  cM is the concentration of the inert molecule

7 0
4 years ago
Assume that 25 ml of 0.440M NaCl are added to 25 ml of 0.320M AgNO3.
Genrish500 [490]
The balanced chemical equation would be as follows:

<span>NaCl + AgNO3 -> NaNO3 + AgCl

We are given the amounts of the reactants. We need to determine first which one is the limiting reactant. We do as follows:

0.0440 mol/L NaCl (.025 L) = 0.0011 mol NaCl -----> consumed completely and therefore the limiting reactant
0.320 mol/L AgNO3 (0.025 L) = 0.008 mol AgNO3

0.0011 mol NaCl ( 1 mol AgCl / 1 mol NaCl) = 0.0011 AgCl precipitate produced


</span>
8 0
3 years ago
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