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3 years ago

The measurement 0.0002833g has how many significant figures?

Chemistry

1 answer:

kherson [118]3 years ago

6 0

Answer:

There are seven significant figures

Explanation:

There are seven different digits within the number. Three 0s, one 2, one 8, and two 3s, adding up to seven different numbers. You exclude the first 0 when the number is a decimal, leaving seven significant figures. Hope this makes sense! :)

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Cellular respiration involves a reaction between glucose and oxygen to form

kodGreya [7K]

From my research and own thoughts I think D

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3 years ago

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the solubility of nitrogen gas is 1.90 mL/dL of blood at 1.00 atm. what is the solubility of nitrogen gas in a deepsea divers bl

leonid [27]

The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.

We want to relate the solubility of a gas with its partial pressure.

We can do so using Henry's law.

<h3>What does Henry's law state?</h3>

Henry's law states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.

C = k × P

where,

C is the concentration of a dissolved gas.

k is the Henry's Law constant.

P partial pressure of the gas.

The solubility of nitrogen gas is 1.90 mL/dL of blood at 1.00 atm.

Since the solvent is basically water, we can understand that the concentration of nitrogen gas is 1.90 mL/dL at 1.00 atm.

We can use this information to calculate Henry's Law constant.

k = C/P = (1.90 mL/dL)/1.00 atm = 1.90 mL/dL.atm

We want to calculate the solubility of nitrogen gas at a pressure of 7.00 atm.

We will use Henry's law.

C = k × P = (1.90 mL/dL.atm) × 7.00 atm = 13.3 mL/dL

The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.

Learn more about solubility here: brainly.com/question/11963573

6 0

2 years ago

328. mL of 0.00345 M NaI (aq) is combined with 703. mL of 0.00802 M Pb(NO3)2 (aq). Determine if a precipitate will form given th

Vsevolod [243]

Answer:

The solution will not form a precipitate.

Explanation:

The Ksp of PbI₂ is:

PbI₂(s) ⇄ 2I⁻(aq) + Pb²⁺(aq)

Ksp = 1.40x10⁻⁸ = [I⁻]²[Pb²⁺] Concentrations in equilibrium

When 328mL of 0.00345M NaI(aq) is combined with 703mL of 0.00802M Pb(NO₃)₂. Molar concentration of I⁻ and Pb²⁺ are:

[I⁻] = 0.00345M × (328mL / (328mL+703mL) = 1.098x10⁻³M

[Pb²⁺] = 0.00802M × (703mL / (328mL+703mL) = 5.469x10⁻³M

Q = [I⁻]²[Pb²⁺] Concentrations not necessary in equilibrium

If Q = Ksp, the solution is saturated, Q > Ksp, the solution will form a precipitate, if Q < Ksp, the solution is not saturated.

Replacing:

Q = [1.098x10⁻³M]²[5.469x10⁻³M] = 6.59x10⁻⁹

As Q < Ksp, the solution is not saturated and will not form a precipitate.

6 0

3 years ago

An aqueous solution containing 10 g of an optically pure compound was diluted to 500 mL with water and was found to have a speci

lapo4ka [179]

Answer:

Optical purity = 76.9231 %

Specific rotation of mixture = - 97.6923 °

Explanation:

The mass of the racemic mixture = 3 g

It means it contains R enantiomer = 1.5 g

S enantiomer = 1.5 g

Amount of Pure R = 10 g

Total R = 11.5 g

Total volume = 500 mL + 500 mL = 1000 mL = 1 L

[R] = 11.5 g/L

[S] = 1.5 g/L

Enantiomeric excess = $\frac {Excess}{Total\ Concentration}\times 100$ = $\frac {11.5-1.5}{11.5+1.5}\times 100$ = 76.9231 %

Optical purity = 76.9231 %

Also,

Optical purity = $\frac {optical\ rotation\ of\ mixture}{optical\ rotation\ of\ pure\ enantiomer}\times 100$

Optical rotation of pure enantiomer = −127 °

$76.9231=\frac {optical\ rotation\ of\ mixture}{-127^0}\times 100$

Specific rotation of mixture = - 97.6923 °

6 0

3 years ago

For the generic equilibrium HA(aq) ⇌ H+(aq) + A−(aq), which of these statements is true? For the generic equilibrium , which of

timama [110]

Answer: The correct statement is if you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase.

Explanation:

Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect generally decreases the solubility of a solute.

Equilibrium reaction of HA and KA follows the equation:

$HA\rightleftharpoons H^{+}(aq.)+A^{-}(aq.)$

$KA\rightleftharpoons K^+(aq.)+A^{-}(aq.)$

According to Le-Chateliers principle, if there is any change in the variables of the reaction, the equilibrium will shift in the direction in order to minimize the effect.

In the equilibrium reactions, $A^-$ ion is getting increased on the product side, so the equilibrium will shift in the direction to minimize this effect, which is in the direction of HA.

Thus, the addition of KA will shift the equilibrium in the left direction.

Equilibrium constant depends on the temperature of the system. It does not have any effect on any change of pH.

pH is defined as the negative logarithm of hydrogen ions present in the solution

If the solution has high hydrogen ion concentration, then the pH will be low.
If the solution has low hydrogen ion concentration, then the pH will be high.

As, the equilibrium is shifting in the left direction, that means concentration of $H^+$ ions are getting decreases. This will increase the pH of the solution.

Hence, the correct statement is if you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase.

8 0

3 years ago