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balandron [24]
2 years ago
6

Sodium and chlorine atoms combine readily because they both tend to lose electrons. True or False?

Chemistry
1 answer:
Ierofanga [76]2 years ago
6 0
False because sodium give electron and chlorine gains so they become ions and because of opposite charges combine and makes bond
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Carnegie Development stages
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Stage 1: 1 days.

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What mass of the protein gelatin is needed to make 0.5 L of a 3 g/L gelatin solution? Show your work.
REY [17]

Answer:

m = 1.5 gram

Explanation:

Given that,

Density of protein gelatin, d = 3 g/L

The volume of protein gelatin, V = 0.5 L

We need to find the mass of the protein gelatin. The density of an object is given by :

d = m/V

Where

m is mass

m=d\times V\\\\m=3\ g/L\times 0.5\ L\\\\m=1.5\ g

So, the required mass is 1.5 gram.

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2 years ago
Elements in the periodic table are listed in order of their -----​
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What is inside of a galactic center? (Inner bulge/bar of a galaxy) What kind of gases does it contain?
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8 0
3 years ago
The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure
Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate \Delta H_{vap} of the reaction, we use clausius claypron equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = vapor pressure at temperature T_1

P_2 = vapor pressure at temperature T_2

\Delta H_{vap} = Enthalpy of vaporization  

R = Gas constant = 8.314 J/mol K

1) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =363 K

T_2 = final temperature =373 K

P_2=1 atm, P_1=?

Putting values in above equation, we get:

\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]

P_1=0.69671 atm \approx 0.6970 atm

The vapor pressure at temperature 363 K is 0.6970 atm

2) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =373 K

T_2 = final temperature =383 K

P_1=1 atm, P_2?

Putting values in above equation, we get:

\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]

P_2=1.4084 atm \approx 1.410 atm

The vapor pressure at 383 K is 1.410 atm

8 0
3 years ago
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