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3 years ago

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The recommended daily allowance of niacin (vitamin B3) is 12 mg per day for children. The molecular formula for niacin is C6H5NO

2. Calculate the number of molecules of niacin in 12 mg .

1 answer:

Tom [10]3 years ago

4 0

Answer:

5.87 ×10¹⁹ molecules.

Explanation:

From the question given above, the following data were obtained:

Mass of C₆H₅NO₂ = 12 mg

Number of molecules of C₆H₅NO₂ =?

Next, we shall convert 12 mg to g. This can be obtained as follow:

1 mg = 10¯³ g

Therefore,

12 mg = 12 mg × 10¯³ g / 1 mg

12 mg = 0.012 g

Next, we shall determine the mass of 1 mole of C₆H₅NO₂. This can be obtained as follow:

1 mole of C₆H₅NO₂ = (12×6) + (5×1) + 14 + (16×2)

= 72 + 5 + 14 + 32

= 123 g

Finally, we shall determine the number of molecules present in 12 mg (i.e 0.012 g) of C₆H₅NO₂. This can be obtained as follow:

From Avogadro's hypothesis,

1 mole of C₆H₅NO₂ = 6.02×10²³ molecules

123 g of C₆H₅NO₂ = 6.02×10²³ molecules

Therefore,

0.012 g of C₆H₅NO₂ = 0.012 × 6.02×10²³ / 123

0.012 g of C₆H₅NO₂ = 5.87 ×10¹⁹ molecules.

Thus, 12 mg (i.e 0.012 g) of C₆H₅NO₂ contains 5.87 ×10¹⁹ molecules.

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2 g/cm³ is the density of the block of wood

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To determine density of the block of wood we use the following formula:

density = mass / volume

density of the wood block = 100 g / 50 cm³

density of the wood block = 2 g/cm³

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3 years ago

Which of the following experimental methods cannot be used to measure the rate of a reaction?

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What mass of sodium chloride contains 4.59 x 10 24 formula units?

shtirl [24]

The term formula units means molecules.

Then, what you are looking for is the mass in 4.59*10^24 molecules.

The procedure involves to convert the 4.59 * 10^24 molecules into moles and use the molar mass of the sodium chloride.

1) Number of moles = 4.59 * 10^24 molecules / (6.02 * 10^23 molecules/mol) = 7.62 mol

2) Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol

3) mass of NaCl = molar mass * number of moles = 58.44 g/mol * 7.62 mol = 445.31 g of NaCl

Answer: 445.31 g of NaCl.

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3 years ago

The coefficient of thermal expansion α = (1/V)(∂V/∂T)p. Using the equation of state, compute the value of α for an ideal gas. Th

andreyandreev [35.5K]

Answer:

The coefficient of thermal expansion α is

$\alpha = \frac{1}{T}$

The coefficient of compressibility

$\beta = \frac{1}{P}$

Now considering $(\frac{ \delta P }{\delta T} )V$

From equation (1) we have that

$\frac{ \delta P}{\delta T} = \frac{n R }{V}$

From ideal equation

$nR = \frac{PV}{T}$

So

$\frac{\delta P}{\delta T} = \frac{PV}{TV}$

=> $\frac{\delta P}{\delta T} = \frac{P}{T}$

=> $\frac{\delta P}{\delta T} = \frac{\alpha }{\beta}$

Explanation:

From the question we are told that

The coefficient of thermal expansion is $\alpha = \frac{1}{V} * (\frac{\delta V}{ \delta P}) P$

The coefficient of compressibility is $\beta = - (\frac{1}{V} ) * (\frac{\delta V}{ \delta P} ) T$

Generally the ideal gas is mathematically represented as

$PV = nRT$

=> $V = \frac{nRT}{P} --- (1)$

differentiating both side with respect to T at constant P

$\frac{\delta V}{\delta T } = \frac{ n R }{P}$

substituting the equation above into $\alpha$

$\alpha = \frac{1}{V} * ( \frac{ n R }{P}) P$

$\alpha = \frac{nR}{PV}$

Recall from ideal gas equation $T = \frac{PV}{nR}$

So

$\alpha = \frac{1}{T}$

Now differentiate equation (1) above with respect to P at constant T

$\frac{\delta V}{ \delta P} = -\frac{nRT}{P^2}$

substituting the above equation into equation of $\beta$

$\beta = - (\frac{1}{V} ) * (-\frac{nRT}{P^2} ) T$

$\beta =\frac{ (\frac{n RT}{PV} )}{P}$

Recall from ideal gas equation that

$\frac{PV}{nRT} = 1$

So

$\beta = \frac{1}{P}$

Now considering $(\frac{ \delta P }{\delta T} )V$

From equation (1) we have that

$\frac{ \delta P}{\delta T} = \frac{n R }{V}$

From ideal equation

$nR = \frac{PV}{T}$

So

$\frac{\delta P}{\delta T} = \frac{PV}{TV}$

=> $\frac{\delta P}{\delta T} = \frac{P}{T}$

=> $\frac{\delta P}{\delta T} = \frac{\alpha }{\beta}$

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3 years ago

What is the pH of a KOH solution that has [H ] = 1. 87 × 10–13 M? What is the pOH of a KOH solution that has [OH− ] = 5. 81 × 10

vlada-n [284]

pH is the hydrogen ion concentration and pOH is the hydroxide ion concentration in the solution. pH KOH is 12.73, pOH KOH is 2.24 and pH NaCl is 7.

<h3>What are pH and pOH?</h3>

pH is the negative log of the hydrogen ion concentration and pOH is the negative log of the hydroxide ion concentration.

The relation between the pH and pOH can be given as, $\rm pOH = 14 - pH$

The pH of KOH can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the first case, the concentration of the KOH is $1. 87 \times 10^{-13}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 87 \times 10^{-13}\;\rm M ]\\\\&= 12.73\end{aligned}$

Hence, the pH of KOH is 12.73.

<u />

pOH of KOH can be calculated by the formula,

$\rm pOH = \rm -log [OH^{-}]$

The hydroxide concentration of the KOH solution is $5. 81 \times 10^{-3}\;\rm M$

Substituting value in the equation:

$\begin{aligned} \rm pOH &= \rm -log [OH^{-}]\\\\&= \rm -log [5. 81 \times 10^{-3}\;\rm M ]\\\\&= 2.24 \end{aligned}$

Hence, the pOH of KOH is 2.24

<u />

The pH of NaCl can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the third case, the concentration of the NaCl is $1. 00\times 10^{-7}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 00 \times 10^{-7}\;\rm M ]\\\\&= 7 \end{aligned}$

Hence, the pH of KOH is 7.0.

Therefore, KOH is basic and NaCl is approximately neutral.

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3 years ago