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10. A small gold nugget has volume of 0.87 cm3. What is its mass if the density of gold is 19.3 g/cm3?

bulgar [2K]

Answer:

16.791 grams

Explanation:

The density formula is:

$d=\frac{m}{v}$

Rearrange the formula for m, the mass. Multiply both sides of the equation by v.

$d*v=\frac{m}{v}*v$

$d*v=m$

The mass of the gold nugget can be found by multiplying the density and volume. The density is 19.3 grams per cubic centimeter and the volume is 0.87 cubic centimeters.

$d= 19.3 g/cm^3\\v-0.87 cm^3$

Substitute the values into the formula.

$m=d*v$

$m= 19.3 g/cm^3*0.87 cm^3$

Multiply. Note that the cubic centimeters, or cm³ will cancel each other out.

$m=19.3 g*0.87$

$m=16.791 g$

The mass of the gold nugget is 16.791 grams.

8 0

3 years ago

In Mendelian genetics, when a tall plant is crossed with a short plant, the offspring will be:

Annette [7]

Some offspring may be tall, some may be short, and some may be medium medium sized.

3 0

3 years ago

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Which of these requires accurate coefficients in a reaction?

aksik [14]

Answer:

A: molar ratio

Molar ratios state the proportions of reactants and products that are used and formed in a chemical reaction.

6 0

2 years ago

Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)

Reptile [31]

<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

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

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

For the given chemical reaction:

$C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)$

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

$\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]$

We are given:

$\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J$

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

4 0

2 years ago

For the reaction shown, identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent. KNO3 →

Vinvika [58]

Answer : The oxidizing element is N and reducing element is O.

$KNO_{3}$ is act as an oxidizing agent as well as reducing agent.

Explanation :

An Oxidizing agent is the agent which has ability to oxidize other or a higher in oxidation number.

Reducing agent is the agent which has ability to reduce other or lower in oxidation number.

The given reaction is :

$KNO_{3} \rightarrow KNO_{2} +O_{2}$

$KNO_{3}$ act as an oxidizing agent.

The oxidation number of N in $KNO_{3}$ is calculated as:

(+1)+(x)+3(-2) = 0

x = +5

And the oxidation number of N in $KNO_{2}$ is calculated as:

(+1)+(x)+2(-2) = 0

x = +3

From the oxidation number method, we conclude that the oxidation number reduced this means $KNO_{3}$ itself get reduced to $KNO_{2}$ and it can act as an oxidizing agent.

$KNO_{3}$ act as a reducing agent.

$KNO_{3} \rightarrow KNO_{2} +O_{2}$

The oxidation number of O in $KNO_{3}$ is calculated as:

(+1)+(+5)+3(x) = 0

x = -2

The oxidation number of O in $O_{2}$ is Zero (o).

Now, we conclude that the oxidation number increases this means $KNO_{3}$ itself get oxidized to $O_{2}$ and it can act as reducing agent.

4 0

3 years ago

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