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torisob [31]
4 years ago
10

Properties of covalent compounds include: (More than 1 may be selected)

Chemistry
1 answer:
denpristay [2]4 years ago
5 0

Answer:

E

C eg In graphites and diamonds

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According to the law of conservation of mass, in all chemical reactions _____.
Fynjy0 [20]

changing the bonds changes the mass.

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7 0
3 years ago
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Dissolved: Will give brainliest
Anna [14]

Answer:

im 99% sure it's 175.

Explanation:

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3 years ago
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Aluminum and cloride undergoes a synthesis reaction if 8molnof Al reacts with 10mol of cl, what is the maximum amount of AlCl3 c
Inessa [10]

<u>Answer:</u> The mass of AlCl_3 produced is 889.38 g

<u>Explanation:</u>

We are given:

Moles of Al = 8 mol

Moles of Cl_2 = 10 mol

For the given chemical reaction:

2Al+3Cl_2\rightarrow 2AlCl_3

By stoichiometry of the reaction:

If 3 moles of chlorine gas reacts with 2 moles of Al

So, 10 moles of chlorine gas will react with = \frac{2}{3}\times 10=6.67mol of Al

As the given amount of Al is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

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

By the stoichiometry of the reaction:

If 3 moles of Cl_2 produces 2 mole of AlCl_3

So, 10 moles of Cl_2 will produce = \frac{2}{3}\times 10=6.67mol of AlCl_3

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

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

We know, molar mass of AlCl_3 = 133.34 g/mol

Putting values in above equation, we get:

\text{Mass of }AlCl_3=(6.67mol\times 133.34g/mol)=889.38g

Hence, the mass of AlCl_3 produced is 889.38 g

4 0
3 years ago
In a coffee-cup calorimeter, 1 mol NaOH and 1 mol HBr initially at 22.5 oC (Celsius) are mixed in 100g of water to yield the fol
zavuch27 [327]

Answer:

ΔH = -55.92 kJ

Explanation:

<u>Step 1:</u> Data given

1 mol NaOH and 1 mol HBr initially at 22.5 °C are mixed in 100g of water

After mixing the temperature rises to 83 °C

Specific heat of the solution = 4.184 J/g °C

Molar mass of NaOH = 40 G/mol

Molar mass of HBr = 80.9 g/mol

<u>Step 2: </u>The balanced equation

NaOH + HBr → Na+(aq) + Br-(aq) + H2O(l)

<u>Step 3:</u> mass of NaOH

Mass = moles * Molar mass

Mass NaOH = 1 * 40 g/mol

Mass NaOH = 40 grams

Step 4: Mass of HBr

Mass HBr = 1 mol * 80.9 g/mol

Mass HBr = 80.9 grams

Step 5: Calculate ΔH

ΔH = m*c*ΔT

ΔH= (100 + 40 + 80.9) * 4.184 * (83-22.5)

ΔH= 220.9 * 4.184 * 60.5

ΔH= 55916.86 J = 55.92 kJ

Since this is an exothermic reaction, the change in enthalpy is negative.

ΔH = -55.92 kJ

4 0
4 years ago
What is the percent mass oxygen in calcium carbonate (CaCo3)?
Gnoma [55]
  <span>Step 1 is to determine the mass of each part 
Mass of Ca is 40.08 g 
Mass of C is 12.01 g 
Mass of O is 16.00 x 3 = 48.00 g 
Step 2 is to determine the total mass of the compound 
Total mass of CaCO3 is 40.08 + 12.01 + 48.00 = 100.09 g 

Step 3 is to determine the % of each part using the following formula: 
Mass of part / total mass x 100 = 

40.08 / 100.09 x 100 = 40.04 % Ca 

12.01 / 100.09 x 100 = 12.00 % C 

48.00 / 100.09 x 100 = 47.96 % O 

Step 4 is to double check by adding all percentages. If they equal 100, then I probably did it right. :) 
40.04 
+12.00 
+47.96 
=100.00</span><span>
</span>
4 0
3 years ago
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