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Nat2105 [25]
3 years ago
15

Calculate the percent error in the atomic weight if the mass of a Zn electrode increased by 0.3681g and 6.514x10-3 moles of Zn w

as produced. You may assume the molar mass of elemental zinc is 65.38 g/mol.
a. -13.5%
b. 13.52%
c. -13.52%
d. 13.5%
Chemistry
1 answer:
evablogger [386]3 years ago
3 0

Answer:

13.5%

Explanation:

To answer this you need to understand that percentage error can never be negative as we use absolute value in the equation

To get molar/atomic mass = mass/moles

=0.3681g/(6.514×10^-3)moles

=56.509g/moles

percentage error = (65.38-56.509)/65.38 ×100

=13.56% (this is what I'm getting without rounding off your question wasn't specific on how many d.p to use)

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Answer:

Equilibrium concentration of H_{2}O is 12.5 M

Explanation:

Given reaction: C_{2}H_{4}+H_{2}O\rightleftharpoons C_{2}H_{5}OH

Here, K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}

where K_{c} represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations

Here, [C_{2}H_{4}]=0.015M , [C_{2}H_{5}OH]=1.69M and K_{c}=9.0

So, [H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M

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The total volume of seawater is 1.5 x 10²¹ L. Seawater contains approximately 3.5% sodium chloride by mass. At that high of a co
garri49 [273]

Answer:

There are 5.408\times 10^{22} grams contained in all the seawater in the world.

Explanation:

At first let is determinate the total mass of seawater (m_{sw}), measured in grams, in the world by definition of density and considering that mass is distributed uniformly:

m_{sw} = \rho_{sw}\cdot V_{sw}

Where:

\rho_{sw} - Density of seawater, measured in grams per liters.

V_{sw} - Volume of seawater, measured in liters.

If V_{sw} = 1.5\times 10^{21}\,L and \rho_{sw} = 1030\,\frac{g}{L}, then:

m_{sw}=\left(1030\,\frac{g}{L} \right)\cdot (1.5\times 10^{21}\,L)

m_{sw} = 1.545\times 10^{24}\,g

The total mass of sodium chloride is determined by the following ratio:

r = \frac{m_{NaCl}}{m_{sw}}

m_{NaCl} = r\cdot m_{sw}

Given that m_{sw} = 1.545\times 10^{24}\,g and r = 0.035, the total mass of sodium chloride in all the seawater in the world is:

m_{NaCl} = 0.035\cdot (1.545\times 10^{24}\,g)

m_{NaCl} = 5.408\times 10^{22}\,g

There are 5.408\times 10^{22} grams contained in all the seawater in the world.

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A bonding that occurs between high electronegative atoms such are N, F, O and H atoms, is called a hydrogen bond. Hydrogen bond is a very strong bond. (C)

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