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Maslowich
3 years ago
5

A cylinder rod formed from silicon is 21.3 cm long and has a mass of 5.00 kg. The density of silicon is 2.33 g/cm3. What is the

diameter of the cylinder? (the volume of cylinder is given by π r2h, where r is the radius and h is the length)
Chemistry
1 answer:
Vinil7 [7]3 years ago
6 0

Answer:

11.32cm

Explanation:

This question describes a cylinder rod formed from silicon with length 21.3cm and mass 5kg. The density of the silicon is 2.33g/cm3.

To calculate the diameter, the radius is needed. To calculate the radius, the volume is needed. To calculate the volume, the formula: density = mass (m) / volume (V) is used.

Mass = 5kg, which is 5 kg × 1000g = 5000g

Density= 2.33g/cm3

Hence; volume= mass / density

= 5000/2.33

= 2145.9 cm3

Volume of cylinder= πr^2h

Where h= 21.3cm and π= 3.142

That is; r^2 = volume/πh

= r^2 = 2145.9/3.142×21.3

= r^2 = 2145.9/66.9246

= r^2 = 32.06

r= √32.06

r= 5.66cm

If radius of the cylinder is 5.66cm, the diameter is twice of the radius.

That is, diameter (d) = 5.66 × 2

= 11.32 cm

Therefore, the diameter of the cylinder is 11.32cm.

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Explanation:

6 0
3 years ago
A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

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pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

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