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3 years ago

9

Two important ways that energy is transported in the world around us is through __ and _.

2 answers:

lana66690 [7]3 years ago

8 0

The two ways that energy can be transferred are by doing work and by heat transfer.

Paul [167]3 years ago

3 0

Work and heat transfer

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The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C

artcher [175]

Explanation:

The given data is as follows.

$P_{1}$ = 100 mm Hg or $\frac{100}{760}atm$ = 0.13157 atm

$T_{1}$ = $1080 ^{o}C$ = (1080 + 273) K = 1357 K

$T_{2}$ = $1220 ^{o}C$ = (1220 + 273) K = 1493 K

$P_{2}$ = 600 mm Hg or $\frac{600}{760}atm$ = 0.7895 atm

R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

$log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}$

$log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]$

$log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]$

0.77815 = $\frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K$

$\Delta H_{vap}$ = $2.219 \times 10^{5}$ J/mol

= $2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}$

= 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

4 0

2 years ago

Besides plants, do other organism get their energy directly from the Sun?<br> yes or no

Contact [7]

yes they doExplanation:

3 0

2 years ago

Read 2 more answers

What is the charge for H before electron share?

just olya [345]

An H atom is made up of a nucleus with a +1 charge, as well as a single electron. Therefore, the only positively charged ion possible has charge +1.

I hope this helps!

8 0

2 years ago

Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu

pochemuha

Answer : The volume of $CO_2$ will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

$HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2$

First we have to calculate the mass of $HCO_3^-$ in tablet.

$\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g$

Now we have to calculate the moles of $HCO_3^-$ .

Molar mass of $HCO_3^-$ = 1 + 12 + 3(16) = 61 g/mole

$\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles$

Now we have to calculate the moles of $CO_2$ .

From the balanced chemical reaction, we conclude that

As, 1 mole of $HCO_3^-$ react to give 1 mole of $CO_2$

So, 0.0202 mole of $HCO_3^-$ react to give 0.0202 mole of $CO_2$

The moles of $CO_2$ = 0.0202 mole

Now we have to calculate the volume of $CO_2$ by using ideal gas equation.

$PV=nRT$

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = $37^oC=273+37=310K$

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

$(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)$

$V=0.51411L=514.11ml$

Therefore, the volume of $CO_2$ will be, 514.11 ml

6 0

2 years ago

How do you convert 134kj to Calories?

Sonja [21]

1 kilo joule = 0.239006 calories
134 kilo joule = 134 x 0.239006
= 32.026804 calories

5 0

3 years ago