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3 years ago

Anyone know ?? Worth 10

Chemistry

2 answers:

Tamiku [17]3 years ago

7 0

just K Y M kid. to be honest none of this is worth it

katen-ka-za [31]3 years ago

3 0

The botanist is measuring growth, so some kind of length, and he is using a ruler. Measuring length is called linear measurement, so it’s that one.

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Describe a group of family on the periodic table

lana66690 [7]

Metals are all in the center of the periodic table and are all made up of metals

6 0

3 years ago

Read 2 more answers

What is the molar concentration of

Leto [7]

Answer:

1.42 M

Explanation:

First calculate the amount of moles.

that's done by dividing the mass with the molecular mass so 660g / 310.18 g/mol = 2.13 mol

Then you can calculate the molarity by dividing the moles with the volume so 2.13 mol / 1.5 l = 1.42 M

(without rounding: 1.418531175 M)

3 0

2 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

what is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0L

Mamont248 [21]

If x=0 then y=0 sorry dude I wish I could
Help but I’m dumb and I just want points but I’d defiantly recommend to look this up on safari or go into school and seek help

4 0

3 years ago

What is the capillary rise of ethanol in a glass tube with a 0.1 mm radius if the surface tension of ethanol is 0.032Jm2 and the

sweet [91]

Answer: The capillary rise(h) in the glass tube is = 0.009m

Explanation:

Using the equation

h = 2Tcosθ/rpg

Given

Contact angle, θ = Zero

h = height of the glass tube=?

T = surface tension = $0.032J/m^2$

r = radius of the tube = 0.1mm =0.0001m

p= density of ethanol = $0.71g/cm^3$

g= $9.8m/s^2$

h = $(2 * 0.032 * cos 0)/( 710*9.8*0.0001)$

h= 0.09m

Therefore the capillary rise in the tube is 0.09m

5 0

3 years ago