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Molodets [167]
3 years ago
6

What volume (mL) of the sweetened tea described in Example 1 contains the same amount of sugar (mol) as 10 mL of the soft drink

in this example
Chemistry
1 answer:
S_A_V [24]3 years ago
5 0

The question is incomplete, the complete question is:

What volume (mL) of the sweetened tea described in Example 3.14 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example. The example is attached below.

<u>Answer:</u> 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink

<u>Explanation:</u>

We first calculate the number of moles of soft drink in a volume of 10 mL

The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}} .....(1)

Taking the concentration of soft drink from the example be = 0.375 M

Volume of solution = 10 mL

Putting values in equation 1, we get:

0.375=\frac{\text{Moles of sugar in soft drink}\times 1000}{10}\\\\\text{Moles of sugar in soft drink}=\frac{0.375\times 10}{1000}=0.00375mol

<u>Calculating volume of sweetened tea:</u>

Moles of sugar = 0.00375 mol

Molarity of sweetened tea = 0.05 M

Putting values in equation 1, we get:

0.05=\frac{0.00375\times 1000}{\text{Volume of sweetened tea}}\\\\\text{Volume of sweetened tea}=\frac{0.00375\times 1000}{0.05}=75mL

Hence, 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink

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Answer:

The stronger acid is HBrO3.

Explanation:

It has an additional oxygen making it more electronegative, in turn making it a stronger acid.

8 0
3 years ago
Which rule for assigning oxidation numbers is correct?
AURORKA [14]

The incorrect rule for assigning oxidation numbers is Hydrogen is usually –1.

Hydrogen is usually +1

<h3>What is oxidation number?</h3>

Oxidation numbers can be defined as that number which is assigned to an element in chemical reaction which represents the number of electrons lost or gained.

So therefore, the incorrect rule for assigning oxidation numbers is Hydrogen is usually –1.

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7 0
1 year 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
3 years ago
For the equation 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O, how many units of NO3 are represented on the products side?
topjm [15]

Answer:

6

Explanation:

3 * 2 =6

5 0
2 years ago
A reaction produces 74.10 g Ca(OH)2 after 56.08 g CaO is added to 36.04 g H2O. How should the difference in the masses of reacta
melomori [17]
Cao +  H2O  ---->Ca(OH)2
Calculate   the  number  of  each   reactant  and  the  moles  of  the  product
that  is
moles = mass/molar mass
The  moles  of  CaO=  56.08g/  56.08g/mol(molar  mass  of  Cao)=  1mole
the  moles  of  water=  36.04 g/18  g/mol=  2.002moles
The   moles  of Ca (OH)2=74.10g/74.093g/mol= 1mole

 The  mass  of differences  of  reactant  and  product  can   be  therefore 
 explained  as 
 1  mole   of  Cao  reacted  completely   with   1  mole   H2O  to  produce  1 mole  of  Ca(OH)2. The  mass  of  water   was  in  excess  while  that  of  CaO  was  limited

3 0
3 years ago
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