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2 years ago

How high did Joseph Kittinger get up into the atmosphere before he jumped?

Chemistry

2 answers:

Trava [24]2 years ago

8 0

I know the answers I do

mamaluj [8]2 years ago

4 0

Answer:

over 50 years ago, joe kittinger rode a balloon 102,800 feet up to the edge of space. then he jumped out.

Explanation:

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Which atom has the least attraction for the electrons in a bond between that atom and an atom of hydrogen?

Helga [31]

Answer:

carbon

Explanation:

tbh im not sure just guessing

8 0

3 years ago

Question 4 (1 point)

Diano4ka-milaya [45]

In an endothermic reaction products are <u>HIGHER </u>than reactants in potential energy and <u>LESS </u>stable.

Explanation:

Energy is input into the reaction in an endothermic reaction. This means the products are of a higher energy level than the reactants. Therefore the reaction increases Gibb's free energy and reduces entropy. Remember in thermodynamic stability involves an increase in entropy and a decrease in Gibbs free energy. Therefore the products are less stable than the reactants. This is why endothermic reactions do not occur spontaneously like exothermic reactions.

6 0

3 years ago

Phosgene (COCl2) is a toxic substance that forms readily from carbon monoxide and chlorine at elevated temperatures: CO(g) + Cl2

lapo4ka [179]

Answer: Concentration of $CO$ = 0.328 M

Concentration of $Cl_2$ = 0.328 M

Concentration of $COCl_2$ = 0.532 M

Explanation:

Moles of $CO$ and $Cl_2$ = 0.430 mole

Volume of solution = 0.500 L

Initial concentration of $CO$ and $Cl_2$ = $\frac{moles}{volume}=\frac{0.430}{0.500}=0.860M$

The given balanced equilibrium reaction is,

$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

Initial conc. 0.860M 0.860M 0

At eqm. conc. (0.860-x) M (0.860-x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[COCl_2]}{[CO][Cl_2]}$

Now put all the given values in this expression, we get :

$4.95=\frac{x}{(0.860-x)^2}$

By solving the term 'x', we get :

x = 0.532 M

Thus, the concentrations of $CO,Cl_2\text{ and }COCl_2$ at equilibrium are :

Concentration of $CO$ = (0.860-x) M =(0.860-0.532) M = 0.328 M

Concentration of $Cl_2$ = (0.860-x) M = (0.860-0.532) M = 0.328 M

Concentration of $COCl_2$ = x M = 0.532 M

3 0

3 years ago

A phosphate buffer is involved in the formation of urine. The developing urine contains H2PO4 and HPO42- in the same concentrati

Katen [24]

Answer:

The ionization equation is

$H_{2}PO_{4}^{-} +H_{2}O$ ⇄ $HPO_{4}^{-2} +H_{3}O^{+}$ (1)

Explanation:

The ionization equation is

$H_{2}PO_{4}^{-} +H_{2}O$ ⇄ $HPO_{4}^{-2} +H_{3}O^{+}$ (1)

As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.

The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,

$Ka = \frac{[HPO_{4}^{-2}] [H_{3}O^{+}]}{[H_{2}PO_{4}^{-}] [H_{2}O]} = 6.2x10^{-8}$

The pKa is

$-Log (Ka) = -Log (6.2x10^{-8}) = 7.2$

The pKa of H2CO3 is 6,35, thus this a stronger acid than H2PO4. The higher the pKa of an acid greater the capacity to donate protons.

In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.

If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1) moves to the left neutralizing the excess proton concentration.

3 0

3 years ago

9. Suppose that 25.0 mL of a gas at 725 mm Hg and 20°C is converted to standard

storchak [24]

Answer:

V₂ = 22.23 mL

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 25 mL

Initial pressure = 725 mmHg (725/760 =0.954 atm)

Initial temperature = 20 °C (20 +273 = 293 K)

Final pressure = standard = 1 atm

Final temperature = standard = 273.15 K

Final volume = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 0.954 atm × 25 mL × 273.15 K / 293 K × 1 atm

V₂ = 6514.63 mL . atm . K / 293 K . atm

V₂ = 22.23 mL

8 0

2 years ago