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Aneli [31]
3 years ago
12

How many grams are in 23.0 moles of carbon (C)?

Chemistry
1 answer:
Gnesinka [82]3 years ago
3 0
276 grams of carbon in 23.0 moles
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Can atoms be created or destroyed during a chemical reaction?
ValentinkaMS [17]

Answer:

No.

Explanation:

During chemical reaction, atomes cannot be created or destroyed, they can only react together to form <em>E</em><em>l</em><em>e</em><em>m</em><em>e</em><em>n</em><em>t</em><em> </em>or <em>C</em><em>o</em><em>m</em><em>p</em><em>o</em><em>u</em><em>n</em><em>d</em><em> </em>at the <em>P</em><em>r</em><em>o</em><em>d</em><em>u</em><em>c</em><em>t</em><em> </em>side.

7 0
3 years ago
9. What is the name of the molecule?
AnnyKZ [126]

Answer:

3–methyl–2–butanol

Explanation:

To name the compound, we must:

1. Identify the functional group.

2. Give the functional group of the compound the lowest possible count.

3. Locate the longest continuous carbon chain. This gives the parent name of the compound.

4. Identify the substituent group attached.

5. Give the substituent group the lowest possible count.

6. Combine the above to get the name of the compound.

Now, let us obtain the name of the compound.

1. The functional group of the compound is Alcohol i.e —OH.

2. The functional group is located at carbon 2.

3. The longest continuous carbon chain is carbon 4 i.e butane. But the presence of the functional group i.e OH will replace the –e in butane with –ol. Therefore, the compound is butanol.

4. The substituent group attached is methyl i.e CH3.

5. The substituent group is located at carbon 3.

6. Therefore, the name of the compound is:

3–methyl–2–butanol.

8 0
3 years ago
2. What if hydrogen were left over?
valentina_108 [34]

Answer: limiting reactant (or limiting reagent): The reactant that determines the amount of product that can be formed in a chemical reaction.

Explanation: Your welcome buddy.

5 0
2 years ago
A sample of food containing 27 g of fat, 48 g of carbohydrates and 20 g of protein is burned in a bomb calorimeter. In a perfect
rosijanka [135]

Answer:

38.3958 °C  

Explanation:

As,

1 gram of carbohydrates on burning gives 4 kilocalories  of energy

1 gram of protein on burning gives 4 kilocalories  of energy

1 gram of fat on burning gives 9 kilocalories of energy

Thus,

27 g of fat on burning gives 9*27 = 243 kilocalories of energy

20 g of protein on burning gives 4*20 = 80 kilocalories  of energy

48 gram of carbohydrates on burning gives 4*48 = 192 kilocalories  of energy

Total energy = 515 kilocalories

Using,

Q=m_{water}\times C_{water}\times (T_f-T_i)

Given: Volume of water = 23 L = 23×10⁻³ m³

Density=\frac{Mass}{Volume}  

Density of water= 1000 kg/m³

So, mass of the water:  

Mass\ of\ water=Density \times {Volume\ of\ water}  

Mass\ of\ water=1000 kg/m^3 \times {0.023\ m^3}  

Mass of water  = 23 kg

Initial temperature = 16°C  

Specific heat of water = 0.9998 kcal/kg°C  

515=23\times 0.9998\times (T_f-16)

Solving for final temperature as:

<u>Final temperature = 38.3958 °C  </u>

8 0
3 years ago
HELPP
slamgirl [31]

Answer:

In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.

Explanation:

tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=

mass compound

mass H

×100%

\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=

mass compound

mass C

×100%

If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=

10.0g compound

2.5g H

×100%=25%

\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=

10.0g compound

7.5g C

×100%=75%

7 0
3 years ago
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