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3 years ago

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The density of gold is 19.3 grams per cubic centimeter. What is the mass of a bar of gold that measures 15.24 cm in length, 10.1

6 cm in width, and 5.08 cm in height?

1 answer:

kumpel [21]3 years ago

4 0

Answer:

40.75

Explanation:

15.24 × 5.08 × 10.16 =786.579072

786.579072/19.3=40.75

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You are holding four identical balloons each containing 10.0g of a different gas. The balloon containing which gas is the larges

Vaselesa [24]

Answer:

Hydrogen, H_2

Explanation:

mass of each gas is 10.0 g

number of mole = mass/ molar mass

number of moles is directly proportional to volume at constant temp and pressure

this implies that the volume is inversely proportional to molar mass. And Among all the gases in periodic table the molar mass of Hydrogen is the least.

molar mass of H2=2 g/mol

Since, H2 has minimum molar mass then for the same mass of the gases Hydrogen will have maximum volume.

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4 years ago

The iupac name this compound

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Answer:

It's <em>HYDROXIDE</em><em> </em>

Explanation:

You do not call it as hydroxide <em>ion</em><em> </em>because ion always have + or -

I hope this helps

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3 years ago

Read 2 more answers

How many atoms are in 34.2 grams of carbon?

Oksana_A [137]

Answer:

6.02*1022

Explanation:

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There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
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First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
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$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

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