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Citrus2011 [14]
3 years ago
5

Which of the following is equal to 0.45 mm3? (B, right?)

Chemistry
1 answer:
Svetradugi [14.3K]3 years ago
6 0
In order to convert mm³ to m³, we use the conversion factor of 1m = 1000 mm. Also, the conversion factor of mm³ to cm³, we use the factor of 1cm = 10 mm.

= (0.45 mm³) x (1 m/ 1000 mm)³ = 4.5 x 10^-10 m³

= (0.45 mm³) x (1 cm/10 mm)³ = 4.5 x 10^-4 cm³

Thus, the answer is letter B. 
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Explanation:

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A student is given the question: "what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume? the density of gold is 19.3
IRINA_888 [86]

<em>The mass of a gold bar  = 1.424 x 10⁴ gram</em>

<em></em>

<h3><em>Further Explanation</em></h3>

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than type

The unit of density can be expressed in g / cm3 or kg / m3

Density formula:

\large{\boxed{{\bold{\rho~=~\frac{m}{V}}}}

ρ = density

m = mass

v = volume

A common example is the water density of 1 gr / cm3

The mass itself is often equated with weight, even though it is different. Mass is the amount of matter in the matter while weight is related to the gravitational force

<em>Known variable</em>

volume gold bar 7.379 × 10⁻⁴ m³

a density of 19.3 g/cm³

<em>Asked</em>

the mass of a gold bar

<em>Answer</em>

volume is known to be 7.379 × 10⁻⁴ m³, then we change it first to units of cm³ adjusting to units of density

7.379 × 10⁻⁴ m³ = 7.379 × 10² cm³

then:

mass = volume x density

mass = 7.379 × 10² cm³ x 19.3 g/cm³

mass = 1.424 x 10⁴ gram

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Keywords: density, mass, volume, a gold bar

3 0
3 years ago
Read 2 more answers
A substance X contains 10 gram of calcium carbonate calculate the number of mole of calcium carbonate present in X ​
Tasya [4]

\LARGE{ \boxed{  \rm{ \red{Required \: answer}}}}

☃️ Chemical formulae ➝ \sf{CaCO_3}

<h3><u>How to find?</u></h3>

For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.

\boxed{ \sf{No. \: of \: moles =  \frac{given \: weight}{molecular \: weight} }}

<h3><u>Solution:</u></h3>

Atomic weight of elements:

Ca = 40

C = 12

O = 16

❍ Molecular weight of \sf{CaCO_3}

= 40 + 12 + 3 × 16

= 52 + 48

= 100 g/mol

❍ Given weight: 10 g

Then, no. of moles,

⇛ No. of moles = 10 g / 100 g mol‐¹

⇛ No. of moles = 0.1 moles

☄ No. of moles of Calcium carbonate in that substance = <u>0.1 moles</u>

<u>━━━━━━━━━━━━━━━━━━━━</u>

3 0
3 years ago
Read 2 more answers
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