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2 years ago

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A 2.50 gram rectangular object has measurements of 22.0 mm, 13.5 mm, and 12.5 mm. What is the object’s density in units of g/mL?

Use significant figures. Do not enter “g/mL” as part of your answer. Do not use scientific notation.

1 answer:

Verdich [7]2 years ago

4 0

Density is equal to the ratio of mass to the volume.

Mathematical expression is given by:

$Density = \frac{mass}{volume}$ (1)

Mass of the rectangular object = 2.50 g

Volume = $22.0 mm\times 13.5 mm\times 12.5 mm$

= 3712.5 $mm^{3}$

Now, convert $mm^{3}$ into mL

1 $mm^{3}$ = 0.001 mL

Thus, volume = 3.7125 mL

Put the values in formula (1)

$Density = \frac{2.50 g}{3.7125 mL}$

= 0.6734 g/mL

In three significant figures, density is equal to 0.673 g/mL

Hence, density is equal to 0.673 g/mL

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Answer : The oxidizing and reducing agents are, $H_2O_2$ and $Fe^{2+}$ .

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The given redox reaction is:

$2H^+(aq)+H_2O_2(aq)+2Fe^{2+}(aq)\rightarrow 2Fe^{3+}(aq)+2H_2O(l)$

The half oxidation-reduction reactions are:

Oxidation reaction : $Fe^{2+}\rightarrow Fe^{3+}+1e^-$

Reduction reaction : $O^-+1e^-\rightarrow O^{2-}$

The oxidation state of oxygen in $H_2O_2$ and $H_2O$ is, (-1) and (-2) respectively.

In this reaction, 'Fe' is oxidized from oxidation (+2) to (+3) and 'O' is reduced from oxidation state (-1) to (-2). Hence, ' $Fe^{2+}$ ' act as a reducing agent and ' $H_2O_2$ ' act as a oxidizing agent.

Thus, the oxidizing and reducing agents are, $H_2O_2$ and $Fe^{2+}$ .

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2 years ago

The Mond process produces pure nickel metal via the thermal decomposition of nickel tetracarbonyl: Ni(CO)4 (l) → Ni (s) + 4CO (g

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<u>Answer:</u> The volume of CO formed is 254.43 L.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of $Ni(CO)_4$ = 444 g

Molar mass of $Ni(CO)_4$ = 170.73 g/mol

Putting values in above equation, we get:

$\text{Moles of }Ni(CO)_4=\frac{444g}{170.73g/mol}=2.60mol$

For the given chemical reaction:

$Ni(CO)_4(l)\rightarrow Ni(s)+4CO(g)$

By stoichiometry of the reaction:

1 mole of nickel tetracarbonyl produces 4 moles of carbon monoxide.

So, 2.60 moles of nickel tetracarbonyl will produce = $\frac{4}{1}\times 2.60=10.4mol$ of carbon monoxide.

Now, to calculate the volume of the gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of the gas = 752 torr

V = Volume of the gas = ? L

n = Number of moles of gas = 10.4 mol

R = Gas constant = $62.364\text{ L Torr }mol^{-1}K^{-1}$

T = Temperature of the gas = $22^oC=(273+22)K=295K$

Putting values in above equation, we get:

$752torr\times V=10.4mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 295K\\\\V=254.43L$

Hence, the volume of CO formed is 254.43 L.

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What orbitals are used to form the 10 sigma bonds in propane (ch3ch2ch3)? Label each atom with the appropriate hybridization. Dr

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Mixing of pure orbitals having nearly equal energy to form equal number of completely new orbitals is said to be hybridization.

For the compound, $CH_3CH_2CH_3$ the electronic configuration of the atoms, carbon and hydrogen are:

Carbon (atomic number=6): In ground state= $1s^{2}2s^{2}2p^{2}$

In excited state: $1s^{2}2s^{1}2p^{3}$

Hydrogen (atomic number=1): $1s^{1}$

All the bonds in the compound is single bond( $\sigma$ -bond) that is they are formed by head on collision of the orbitals.

The structure of the compound is shown in the image.

The Carbon-Hydrogen bond is formed by overlapping of s-orbital of hydrogen to p-orbital of carbon.

In order to complete the octet the required number of electrons for carbon is 4 and for hydrogen is 1. So, the electron in $1s^{1}$ of hydrogen will overlap to the 2p^{3}-orbital of carbon.

Thus, the hybridization of Hydrogen is $s$ -hybridization and the hybridization of Carbon is $sp^{3}$ -hybridization.

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<h2>Answer:</h2>

1.58 × 10∧16 pools.

<h3>Explanation:</h3>

Given:

Length of ice berg= 98 miles = 1557716 meters

Width of iceberg = 25 miles = 40233.6 meters

Thickness of iceberg = 750 ft = 230 meters

Volume of water in a swimming pool = 24,000 gallons = 90850 liters

The volume of the ice berg:

Volume = Length . width . thickness

Volume = 1557716 . 40233.6 . 230 = 1,441, 468, 016, 5248 m3 = 1,441, 468, 016, 5248 × 10 ∧3 L.

1 pool contains liters of water: 90850 liters

1,441, 468, 016, 5248 × 10 ∧3 liters contains = 1/90850 . 1,441, 468, 016, 524.8 × 10 ∧3 .

= 1.100 . 1,441, 468, 016, 5248 × 10 ∧3 L.

= 1.58 × 10∧16 pools.

Hence 1.6 × 10∧16 pools will be filled with that chunk of ice.

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