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3 years ago

Round to 4 significant figures. 35.5450

Chemistry

2 answers:

sineoko [7]3 years ago

8 0

Answer:

35.5450 to 4 significant figures is 35.55

Hope this helps.

Masja [62]3 years ago

3 0

Answer:

35.5450 will be rounded to 35.55

Explanation:

=35.5450

if the last digit is less than 5 then it will be ignored

=35.545

when the dropping digit is 5 then the retaining digit will increse by a factor of 1

=35.55

i hope this will help you

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How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and

Kay [80]

Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

$(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)$

The expression used will be:

$Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]$

where,

$Q$ = heat released for the reaction = ?

m = mass of benzene = 94.4 g

$c_{p,s}$ = specific heat of solid benzene = $1.51J/g^oC=1.51J/g.K$

$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC=1.73J/g.K$

$\Delta H_{fusion}$ = enthalpy change for fusion = $-9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g$

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

$Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]$

$Q=-29427.312J=-29.4kJ$

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

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3 years ago

24. What volume of a 0.0200M calcium hydroxide is required to neutralize 35.00 mL of 0.0500M nitric acid

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Answer:

THE VOLUME OF 0.200M CALCIUM HYDROXIDE NEEDED TO NEUTRALIZE 35 mL of 0.050 M NITRIC ACID IS 43.75 mL.

Explanation:

Using

Ca VA / Cb Vb = Na / Nb

Ca = 0.0500 M

Va = 35 mL

Cb = 0.0200 M

Vb = unknown

Na = 2

Nb = 1

Equation for the reaction:

Ca(OH)2 + 2HNO3 --------> Ca(NO3)2 + 2H2O

So therefore, we make Vb the subject of the equation and solve for it

Vb = Ca Va Nb / Cb Na

Vb = 0.0500 * 35 * 1 / 0.0200 * 2

Vb = 1.75 / 0.04

Vb = 43.75 mL

The volume of 0.02M calcium hydroxide required to neutralize 35 mL of 0.05 M nitric acid is 43.75 mL

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3 years ago

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2 years ago

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3 years ago

Consider an amphoteric hydroxide, m(oh)2(s), where m is a generic metal. estimate the solubility of m(oh)2 in a solution buffere

Colt1911 [192]

Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7
∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
= (2x10^-16)/(1x10^-7)^2
= 0.02 M
b) at PH =10
when POH = 14- PH = 14-10 = 4
when POH = -㏒[OH-]
    4 = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
= 2x10^-16 / (1x10^-4)^2
  = 2x10^-8 M
c) at PH= 14
when POH = 14-PH
= 14 - 14
= 0
when POH = -㏒[OH]
  0 = - ㏒[OH]
∴[OH] = 1 m
by substitution with this value in Ksp formula :
[M2+] = Ksp / [OH]^2
= (2x10^-16) / 1^2
= 2x10^-16 M

8 0

3 years ago