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motikmotik
2 years ago
7

If the results of an experiment disprove a hypothesis, then the a. results should not be reported. b. hypothesis is just a theor

y. c. data must contain errors. d. none of the above
Chemistry
1 answer:
Alina [70]2 years ago
3 0
B.Hypothesis is just a theory.
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Suppose that 25.0 mL of 0.10 M CH3COOH (aq) is titrated with 0.10 M NaOH (aq). What is the pH after the addition of 10.0 mL of 0
olya-2409 [2.1K]

Answer:

pH=-1.37

Explanation:

We are given that 25 mL of 0.10 M CH_3COOH is titrated with 0.10 M NaOH(aq).

We have to find the pH of solution

Volume of CH_3COOH=25mL=0.025 L

Volume of NaoH=0.01 L

Volume of solution =25 +10=35 mL=\frac{35}{1000}=0.035 L

Because 1 L=1000 mL

Molarity of NaOH=Concentration OH-=0.10M

Concentration of H+= Molarity of CH_3COOH=0.10 M

Number of moles of H+=Molarity multiply by volume of given acid

Number of moles of H+=0.10\times 0.025=0.0025 moles

Number of moles of OH^-=0.10\times 0.01=0.001mole

Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles

Concentration of H+=\frac{0.0015}{0.035}=4.28\times 10^{-2} m/L

pH=-log [H+]=-log [4.28\times 10^{-2}]=-log4.28+2 log 10=-0.631+2

pH=-1.37

6 0
3 years ago
HELP PLEASE I NEED HELP THANKS I LOVE U
Pani-rosa [81]

Answer:

0.500-Molarity solution

Explanation:

4 0
3 years ago
The combination of potassium-sparing diuretics and salt substitutes can result in dangerously high blood levels of:
alisha [4.7K]

Answer:

b. potassium.  

Explanation:

Potassium-sparing diuretics and salt substitutes are diuretics that eliminate salt and water but save potassium. They act by inhibiting the conducting sodium channels in the collecting tubule, such as amiloride and triamterene, or by blocking aldosterone, such as spironolactone.

Concomitant use of potassium-sparing diuretics together with salt substitutes may result in dangerously high blood levels of serum potassium. For this reason, it is important to consult a physician before taking these substances at the same time to avoid potential problems with potassium accumulation.

4 0
3 years ago
How many grams of lithium sulfide must be dissolved in 1600.0 g water to make a 2.0 molal solution?
klasskru [66]
Molality is the number of moles of solutes in 1 kg of solvent.
the molality of solution to be prepared is 2.0 molal. 
therefore 2 moles in 1 kg water.
the mass of Li₂S required is - 46 g/mol x 2.0 mol = 92 g
the mass in 1 kg of solvent is - 92 g
Therefore mass of Li₂S required in 1600.0 g is - 92 g/kg x 1.6 kg = 147.2 g
7 0
2 years ago
When an acid reacts with a strong base which product always forms
patriot [66]

Answer:طيزي

Explanation:

4 0
2 years ago
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