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4 years ago

7

A sample of metal has a mass of 18.20g, and a volume of 7.38mL. What is the density in g/cm^3?

1 answer:

AlexFokin [52]4 years ago

8 0

Density = Mass ÷ Volume. You get 2.466

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How do chemists solve problems?

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Answer:Chemistry problems can be solved using a variety of techniques.

Explanation: Many chemistry teachers and most introductory chemistry texts illustrate problem solutions using the factor-label method. ... The use of analogies and schematic diagrams results in higher achievement on problems involving moles, stoichiometry, and molarity. Hope this helped!

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3 years ago

Water is believed to be one of the best fire extinguishers then why it is so that fire in oil spills over ocean/sea is difficult

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3 years ago

Position vs Time

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The starting position of this object is 3 m.

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3 years ago

A group of students is investigating whether copper is a better thermal conductor than aluminum. The students take a copper wire

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3 years ago

Read 2 more answers

One liter of a buffer contains 2.00 Molar NaA and 2.00 Molar HA. (A- is the anion of an acid) 15.00 g NaOH is added. (Assume no

madam [21]

Answer:

The new pH is 10.8031

Explanation:

Given information:

[NaA] = concentration = 2 M

[HA] = concentration = 2 M

V = volume = 1 L

15 g of NaOH added

Question: What is the new pH, pH = ?

Moles of NaA:

$n_{NaA} =2\frac{moles}{L} *1L=2moles$

Moles of HA:

$n_{HA} =2\frac{moles}{L} *1L=2moles$

Moles of NaOH:

$n_{NaOH} =15g*\frac{1mol}{40g} =0.375moles$

The reaction:

HA + NaOH → NaA + H₂O

Moles of NaA = 2 + 0.375 = 2.375 moles

Moles of HA = 2 - 0.375 = 1.625 moles

The value of Ka is 2.3x10⁻¹¹

The pKa:

$pKa=-logKa=-log(2.3x10^{-11} )=10.6383$

Applying the Henderson's Hasselbach equation

$pH=pKa+log\frac{molesNaA}{molesHA} =10.6383+log\frac{2.375}{1.625} =10.8031$

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3 years ago