Chemistry, 05.03.2020 03:43 haileyhale5

A volume of 90.0 mL mL of a 0.590 M M HN O 3 HNO3 solution is titrated with 0.350 M M KOH KOH . Calculate the volume of KOH KOH required to reach the equivalence point. Express your answer to three significant figures and include the appropriate units.

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Answer from: lizdeleon248

152 mL is the volume of KOH required to reach the equivalence point.

Explanation:

$HNO_3(aq)+KOH(aq)\rightarrow KNO_3(aq)+H_2O(l)$

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HNO_3

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=1\\M_1=0.590 M\\V_1=90.0 mL\\n_2=1\\M_2=0.350 M\\V_2=?$

Putting values in above equation, we get:

$1\times 0.590 M\times 90.00=1\times 0.350 M\times V_2$

$V_2=\frac{1\times 0.590 M\times 90.0 mL}{1\times 0.350 M}=151 .7 mL\approx 152 mL$

152 mL is the volume of KOH required to reach the equivalence point.

Answer from: lizdeleon248

152 ml.

Explanation:

Given:

Volume of HNO3 = 90 ml

Molar concentration of HNO3 = 0.59 M

Molar concentration of KOH = 0.35 M

Equation of the reaction

KOH + HNO3 --> KNO3 + H2O

Number of moles of HNO3 = molar concentration × volume

= 0.59 × 0.09

= 0.0531 moles.

By stoichiometry, 1 mole of HNO3 reacts with 1 mole of KOH. Therefore,

Number of moles of KOH = 0.0531 moles.

Volume = 0.0531 ÷ 0.350

= 0.152 l

= 152 ml.

Answer from: Quest

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

Answer from: Quest

$\frac{239}{94}pu+\frac{1}{0}n ====> \frac{y}{z}x+\frac{134}{54}xe+3\frac{1}{0}$

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137 subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54 subtract 54 from both sides.

40 = z

answer

$\frac{103}{40}x=\frac{103}{40}zr$

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A volume of 90.0 mL mL of a 0.590 M M HN O 3 HNO3 solution is titrated with 0.350 M M KOH KOH . Calc...

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