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2 years ago

Chemistry help!

Chemistry

1 answer:

harkovskaia [24]2 years ago

4 0

Answer:

Use the formula below

Explanation:

use the moles ratio for this by writing down the reaction and balancing the equation

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How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of alu

Sergio039 [100]

Answer:

297

Explanation:

55-22=33

Then use 33 *10*0.9

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3 years ago

Determine the number of bonding electrons and the number of nonbonding electrons in the structure of SI2. Enter the number of bo

mariarad [96]

Answer:

4, 16,

Explanation:

SI2 is sulphur diiodide. Sulphur is in group sixteen (six valence electrons) while iodine is in group 17(seven valence electrons).

Since there are two iodine atoms and one sulphur atom, the molecule has twenty valence electrons. Out of these twenty valence electrons, only four are bonding electrons. The other sixteen electrons include the four nonbonding electrons found on sulphur and the twelve non bonding electrons found on the two iodine atoms having six nonbonding electrons each.

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3 years ago

Which compound contains a radical

slamgirl [31]

The compound that contains a radical is called NaOH.

Explanation:

Radical, also called Free Radical, in chemistry, a fragment that contains at trivial one unpaired electron. Most molecules include even numbers of electrons, and the covalent chemical bonds including the atoms commonly within a molecule normally consist of pairs of electrons jointly shared by the atoms combined by the bond.

6 0

3 years ago

Read 2 more answers

In a titration of 47.41 mL of 0.3764 M ammonia with 0.3838 M aqueous nitric acid, what is the pH of the solution when 47.41 mL +

Volgvan

Answer: The pH of the solution is 1.136

Explanation:

To calculate the moles from molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

For ammonia:

Molarity of ammonia = 0.3764 M

Volume of ammonia = 47.41 mL = 0.04741 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol$

For nitric acid:

Molarity of nitric acid = 0.3838 M

Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L

Putting values in above equation, we get:

$0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol$

After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol

For the reaction of ammonia with nitric acid, the equation follows:

$NH_3+HNO_3\rightarrow NH_4NO_3$

At $t=0$ 0.0178 0.022

Completion 0 0.0042 0.0178

As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.

To calculate the pH of the reaction, we use the equation:

$pH=-\log[H^+]$