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timofeeve [1]
2 years ago
8

A crystallographer measures the horizontal spacing between molecules in a crystal. The spacing is

Chemistry
2 answers:
erastova [34]2 years ago
7 0

Answer: The total width of a crystal is 1.65 mm.

Explanation:

Horizontal length between the two molecules = 16.5 nm

Width of the 10^5 molecules :16.5 nm\times 10^5=16.5\times 10^5 nm

1 nm=10^{-6} mm

The total width of a crystal in millimeter=16.5\times 10^5\times 10^{-6} mm=1.65 mm

The total width of a crystal is 1.65 mm.

Len [333]2 years ago
7 0

1.65 mm

<h3>Further explanation</h3>

<u>Given:</u>

A crystallographer measures the horizontal spacing between molecules in a crystal. The spacing is  16.5  nm.

<u>Question:</u>

What is the total width of a crystal in millimeters  that is 10⁵  molecules across?

<u>The Process:</u>

We will solve the problem of unit conversion, especially the unit of length. This time we will convert nanometers to millimeters.

Recall that,

\boxed{ \ 1 \ nm = 10^{-9} \ m \ }

\boxed{ \ 1 \ mm = 10^{-3} \ m \ }

Crystallographers observe 10⁵ molecules across with the intermolecular distance in the crystal is 16.5 nm.

\boxed{ \ 10^5 \ molecules \times \frac{16.5 \ nm}{molecules} \times \frac{10^{-9} \ m}{1 \ nm} \times \frac{1 \ mm}{10^{-3} \ m} =? \ }

We cross out the same units.

\boxed{ \ = 10^5 \times 16.5 \times 10^{-9} \times \frac{1 \ mm}{10^{-3}} \ }

\boxed{ \ = 16.5 \times 10^{5 - 9 + 3} \ mm \ }

\boxed{ \ = 16.5 \times 10^{-1} \ mm \ }

\boxed{ \ = \frac{165}{10} \times \frac{1}{10} \ mm \ }

\boxed{ \ = \frac{165}{100} \ mm \ } \rightarrow \boxed{\boxed{ \ 1.65 \ mm \ }}

Thus, the total width of a crystal in millimeters is 1.65 mm.

<h3>Learn more</h3>
  1. What is the speed in km/hr and mi/min (miles/minute)?  brainly.com/question/2088710#
  2. How many milliliters of coffee were in each cup? brainly.com/question/872847
  3. Convert the hippopotamus's weight to kilograms brainly.com/question/2151424

Keywords: crystallographer, measures, the horizontal spacing, between, molecules, in a crystal, 16.5 nm, the total width, a crystal, in millimeters, nanometers, 10^5  molecules across, write, your answer, as a decimal, conversion, the unit of length, 1.65

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Answer:- Volume of the gas in the flask after the reaction is 156.0 L.

Solution:-  The balanced equation for the combustion of ethane is:

2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.

Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

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let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

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Since ethane is present in excess, the remaining volume of it would also be present in the flask.

Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

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The estimated feed rate of logs is 14.3 logs/min.

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That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.

Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.

To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

V=V_c+V_w\\\\M/\rho=M_c/\rho_c+Mw/\rho_w\\\\M/\rho=0.55*M/\rho_c+0.45*M/\rho_w\\\\1/\rho=0.55/\rho_c +0.45/\rho_w\\\\0.55/\rho_c=1/\rho-0.45/\rho_w\\\\0.55/\rho_c=1/(0.64*\rho_w)-0.45/\rho_w=(1/\rho_w)*(\frac{1}{0.64}-\frac{0.45}{1}  )\\\\0.55/\rho_c=1.1125/\rho_w\\\\\rho_c=\frac{0.55}{1.1125}*\rho_w= 0.494*\rho_w

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n=\frac{supply}{M_{log}}=\frac{3617 tons/day}{387.5 lbm}*\frac{2204lbm}{1ton}\\\\n=20573 logs/day=14.3 logs/min

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Now, to calculate the volume of the gas, we use ideal gas equation, which is:

PV = nRT

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