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2 years ago

5

During laparoscopic surgery , carbon dioxide gas is used to expand the abdomen to help create a larger working space. If 4.80 L

of CO 2 gas at 18 °C at 785 mmHg is used, what is the final volume , in liters, of the gas at 37 degrees * C and a pressure of 745 mmHg if the amount of CO 2 does not change ?

1 answer:

Studentka2010 [4]2 years ago

7 0

Answer:

5.37 L

Explanation:

To solve this problem we need to use the PV=nRT equation.

First we <u>calculate the amount of CO₂</u>, using the initial given conditions for P, V and T:

P = 785 mmHg ⇒ 785/760 = 1.03 atm

V = 4.80 L

T = 18 °C ⇒ 18 + 273.16 = 291.16 K

1.03 atm * 4.80 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 291.16 K

We <u>solve for n</u>:

n = 0.207 mol

Then we use that value of n for another PV=nRT equation, where T=37 °C (310.16K) and P = 745 mmHg (0.98 atm).

0.98 atm * V = 0.207 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 310.16 K

And we <u>solve for V</u>:

V = 5.37 L

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During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

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3 years ago

Ava, Inc., issued 7% bonds, dated January 1, with a face amount of $117,200 on January 1, 2016 for an issue price of 106.5. The

natta225 [31]

Complete Question

The complete question is shown on the first uploaded image

Answer:

10a The interest payment on the bond is = $ 8,204

10b The debit to cash for the bond proceeds is = $ 302,000

Explanation:

The explanation is shown on the second uploaded image

8 0

3 years ago

Water vapor is not usually seen.

xeze [42]

Answer:

True

Explanation:

Water vapor is totally invisible.

Without it, there would be no clouds or rain or snow, since all of these require water vapor in order to form. All of the water vapor that evaporates from the surface of the Earth eventually returns as precipitation - rain or snow.

Helped by the <u><em>One</em></u> & <u><em>Only</em></u> <u><em>#QUEEN</em></u> aka <u><em>#DRIPPQUEENMO </em></u>

<u><em>Hoped this helped! :)</em></u>

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2 years ago

Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e

igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em> by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

Explanation:

1) Moles of NaCl , $n_1=1 mol$

Mass of water = m= 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute(NaCl)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p=17.19 Torr$

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose , $n_1=1mol$

Mass of water = m = 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p'$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute ( glucose)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p'=17.19 Torr$

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

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3 years ago

What tool is used in an explosion site to chemically detect explosive residues on objects?

qaws [65]

Answer:

ion mobility spectrometer

Explanation:

it is an handheld chemical dectection device

Hope this helps

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