Which type of bond is present within a water molecule

Chemistry

1 answer:

krok68 [10]3 years ago

3 0

Answer:

Covalent bonds

Explanation:

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500.0 ml of 0.110 m naoh is added to 535 ml of 0.250 m weak acid (ka = 6.37 × 10-5. what is the ph of the resulting buffer?

sattari [20]

The main formulas are pH=pKa + log(Base/Acid) and pKa = -log(Ka)
so firstly, we must find the value of pKa,
Ka=6.37 x 10 ^-5, and then logKa= log (6.37 . 10^-5)= -9.66, so -logKa= +9.66=Ka
next let's find log(Base/Acid)
for that the concentration of NaOH is [NaOH] = 500.0 x 0.110 / 500+535 =0.053M, the concentration of the Acid is [Acid] =535*0.25 / 500+535 =0.12M, so its difference is [Acid]-[NaOH] = 0.12-0.053=0.07
so pH=pKa + log(Base/Acid)= 9.66 + log(0.053 / 0.07)= 9.66-0.36=9.29 
so pH=9.29.

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3 years ago

A doctor orders 2.0 mg of morphine. The vial of morphine on hand is 40. mg per 4.0 mL. How many milliliters of morphine should y

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The correct answer would be 2.0 ml

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3 years ago

18 ) Liquid A and Liquid B are clear liquids. They are placed in open containers and allowed to evaporate. When evaporation is c

arsen [322]

Answer:

Liquid A is a homogeneous mixture and Liquid B is a heterogeneous mixture

Explanation:

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3 years ago

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3 years ago

How many grams of nitric acid HNO₃, are required to neutralize (completely react with) 4.30 grams of Ca(OH)2 according to the ac

Brrunno [24]

Answer:

7.32g of HNO3 are required.

Explanation:

1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.

From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.

2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:

• starting with the 4.30 grams of Ca(OH)2.

• using the molar mass of Ca(OH)2 (74g/mol).

• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .

• using the molar mass of HNO3 (63.02g/mol).

$4.30g\text{ Ca\lparen OH\rparen}_2*\frac{1\text{ mol Ca\lparen OH\rparen}_2}{74g\text{ Ca\lparen OH\rparen}_2}*\frac{2\text{ moles HNO}_3}{1\text{ mole Ca\lparen OH\rparen}_2}*\frac{63.02g\text{ HNO}_3}{1\text{ mole HNO}_3}=7.32g\text{ HNO}_3$

So, 7.32g of HNO3 are required.

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1 year ago