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3 years ago

Some chemical bonds are considered molecules while others are considered compounds.

1 answer:

5 0

Molecule is the general term used to describe any atoms that are connected by chemical bonds. Every combination of atoms is a molecule. A compound is a molecule made of atoms from different elements. All compounds are molecules, but not all molecules are compounds.

One example of a molecule is $Cl_2$ (chlorine).

Examples of a compound is NaCl (sodium chloride) or $H_2O\\$ (water)

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The gravitational force of Earth is causing volcanoes on the moon.

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This is true, the force can bring land up at ease

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4 years ago

A 50.00-g sample of metal at 78.0°C is dropped into cold water. If the metal sample cools to 17.0°C and the specific heat of met

Luba_88 [7]

Answer:−329.4 calorie

Explanation:Heat released by metal = q=m_mc_m(T_2-T1)q=m

−T1)

m_m=50\ g\\c_m=0.108\ cal\ g^{-1}\ \degree C^{-1}\\q=50\times0.108(17-78)=-329.4\ caloriem

=50 g

=0.108 cal g

−1

°C

−1

q=50×0.108(17−78)=−329.4 calorie

Here minus sign indicate that heat is released

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4 years ago

Read 2 more answers

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Alex17521 [72]

Answer:

What changes do you notice? The white you see have undergone coral bleaching. At high temperatures, corals may lose their zooxanthellae, causing corals to lose their color and their main source of food. Once bleaching occurs, the coral colony usually dies.

Explanation:

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2 years ago

In a chemical reaction, the reactant undergoes:

Step2247 [10]

Its B chemical change

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4 years ago

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

4 years ago