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Alex17521 [72]
3 years ago
12

How many capsules containing 75mg of Tamiflu could be produced from 155g of star anise.?

Chemistry
1 answer:
jeyben [28]3 years ago
7 0
From the question you will find that:
one capsule of tamiflu is obtained from 2.6 g of star anise.
1 capsule   = 2.6 g tamiflu
? capsules = 155 g tamiflu
by cross multiplication = \frac{(1 x 155)}{2.6} = 59 capsules
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Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o
Gnom [1K]

Answer: The empirical formula for the given compound is C_5H_7N

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:

C_xH_yN_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.

We are given:

Mass of CO_2=21.363mg=21.363\times 10^3g=21363g

Mass of H_2O=6.125g=6.125\times 10^3g=6125g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 21363 g of carbon dioxide, \frac{12}{44}\times 21363=5826.27g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, \frac{2}{18}\times 6125=680.55 of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = \frac{485.52}{97.73}=4.96\approx 5

For Hydrogen  = \frac{680.55}{97.73}=6.96\approx 7

For Nitrogen = \frac{97.73}{97.73}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is C_5H_7N_1=C_5H_7N

7 0
3 years ago
Does the motion of the moon affects how we see it from Earth? *
irina [24]
Yes that’s why we see it in different shapes all the time
7 0
2 years ago
Which object is a good conductor of heat and electricity?
Alex

Answer:

C

Explanation:

Metals are good conductors of heat

8 0
2 years ago
1.5mol C3H8 from C3H8+5O2-->3CO2+4H2O .how many grams of carbon dioxide are produced
koban [17]

Answer:

\large \boxed{\text{200 g CO}_{{2}}}

Explanation:

We will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

Mᵣ:                                 44.01

            C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

n/mol:    1.5

1. Calculate the moles of CO₂

The molar ratio is 3 mol  CO₂:1 mol C₃H₈

\rm  \text{Moles of CO}_{2} = \text{1.5 mol C$_{3}$H}_{8} \times \dfrac{\text{3 mol CO}_{2}}{\text{1 mol C$_{3}$H}_{8}} =\text{4.5 mol CO}_{2}

2. Calculate the mass of CO₂.

\text{Mass of CO}_{2} = \text{4.5 mol CO}_{2}  \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO$_{2}$}} = \textbf{200 g CO}_{\mathbf{2}}\\\text{The reaction will form $\large \boxed{\textbf{200 g CO}_{\mathbf{2}}}$}

3 0
2 years ago
Simon has collected three samples from the coral reef where he observes marine life. He must determine whether each one is a pur
zysi [14]
SAMPLE A - <span>pure substance.
</span>SAMPLE B - <span>homogeneous mixture.
</span>SAMPLE C - <span>heterogeneous mixture.
</span>Pure substance - <span>constant composition and properties.</span>
Homogeneous mixture - same uniform appearance and composition.
Heterogeneous mixture - <span>not </span>uniform<span> in composition, two phases (liquid and dust).
</span>
7 0
3 years ago
Read 2 more answers
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