What is radium used for

You fall off your skateboard and skin your knee so you wash it and put a bandage on it. Which seven systems are you using most d

Delicious77 [7]

Bodily systems or like other systems?

8 0

3 years ago

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

3 years ago

Which sub-layer thins out into space where there is no air

bija089 [108]

exosphere - 
the outermost region of a planet's atmosphere.

8 0

3 years ago

Read 2 more answers

How many grams are in 7.9*10^-1 moles of argon

Dafna11 [192]

For moles to grams would be the mole which is 7.9*10^-1 times the molar mass of argon

3 0

4 years ago

Average atomic masses listed by IUPAC are based on a study of experimental results. Bromine has two isotopes 73 br and81 br, who

ddd [48]

Answer: The average atomic mass of element bromine is 80.4104 amu.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For _{35}^{79}\textrm{Br}[/tex] isotope:

Mass of $_{35}^{79}\textrm{Br}$ isotope = 78.9183 amu

Percentage abundance of $_{35}^{79}\textrm{Br}$ isotope = 50.69 %

Fractional abundance of $_{35}^{79}\textrm{Br}$ isotope = 0.5069

For $_{35}^{81}\textrm{Br}$ isotope:

Mass of $_{35}^{81}\textrm{Br}$ isotope = 80.9163 amu

Percentage abundance of $_{35}^{81}\textrm{Br}$ isotope = 49.31 %

Fractional abundance of $_{35}^{81}\textrm{Br}$ isotope = 0.4931

Putting values in equation 1, we get:

$\text{Average atomic mass of Bromine}=[(78.9183\times 0.5069)+(80.9163\times 0.4931)]$

$\text{Average atomic mass of Bromine}=80.4104amu$

Hence, the average atomic mass of element bromine is 80.4104 amu.

5 0

3 years ago