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3 years ago

A solid object has a mass of 30 grams and a volume of 10 cm3.

2 answers:

german3 years ago

7 0

To find the density of an object, you need to use the equation: D = M/V. This is density equals mass over volume.

Mass: 30 g
Volume: 10 cm3

30/10 = 3

Density: 3 g/cm3

tresset_1 [31]3 years ago

6 0

Answer: The density of the object is equal to $3g/cm^3$

Explanation:

Density of an object is defined as the ratio of its mass and volume. The chemical equation representing density of an object is:

$\text{Density of an object}=\frac{\text{Mass of an object}}{\text{Volume of an object}}$

We are given:

Mass of an object = 30 grams

Volume of an object = $10cm^3$

Putting values in above equation, we get:

$\text{Density of an object}=\frac{30g}{10cm^3}=3g/cm^3$

Hence, the density of the object is equal to $3g/cm^3$

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Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

zysi [14]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$N_2+2O_2\rightarrow 2NO_2$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = $\frac{121.77}{1}\times 1.90=231.36 J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

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3 years ago

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