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N76 [4]
3 years ago
14

What is the valency of aluminium and nitrate in Al (NO3)3​

Chemistry
1 answer:
VikaD [51]3 years ago
6 0

Answer:

Aluminum loses three valence shell electrons to become the aluminum ion Al+3. Nitrate is a polyatomic ion with the formula NO3 -1.

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What is a disadvantage of using solar energy?
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Answer:

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Explanation:

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2 years ago
Read 2 more answers
(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g
Anarel [89]

Answer:

Part A

 The volume of the gaseous product  is  V = 787L

Part B

The volume of the the engine’s gaseous exhaust is  V_e = 2178 \ L

Explanation:

Part A

From the question we are told that

    The temperature is  T = 350^oC = 350 +273 =623K

     The pressure is  P = 735 \ torr = \frac{735}{760} =  0.967\ atm

     The of  C_8 H_{18} = 100.0g

The chemical equation for this combustion is

               2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}

 The number of moles of  C_8 H_{18} that reacted is mathematically represented as

               n = \frac{mass \ of \  C_8H_{18}  }{Molar \  mass \ of  C_8H_{18} }

The molar mass of  C_8 H_{18} is constant value which is

                  M = 114.23 \ g/mole  

So          n = \frac{100  }{114.23} }

             n = 0.8754 \ moles

The gaseous product in the reaction is CO_2_{(g)} and water vapour

Now from the reaction

    2 moles of C_8 H_{18}  will react with 25 moles of O_2 to give (16 + 18) moles of CO_2_{(g)} and  H_2 O_{(g)}

So

    1 mole of C_8 H_{18} will  react with 12.5 moles of  O_2 to give 17 moles of CO_2_{(g)} and  H_2 O_{(g)}

This implies that

    0.8754 moles of C_8 H_{18} will react with (12.5 * 0.8754 ) moles of O_2 to give  (17 * 0.8754) of CO_2_{(g)} and  H_2 O_{(g)}

So the no of moles of gaseous product is

         N_g = 17 * 0.8754

         N_g = 14.88 \ moles

From the ideal gas law

       PV = N_gRT

making V the subject

        V = \frac{N_gRT}{P}

Where R is the gas constant with a value R = 0.08206 \  L\cdot atm /K \cdot mole

Substituting values

          V = \frac{14.88* 0.08206 *623}{0.967}

          V = 787L

Part B

From the reaction the number of moles of oxygen that reacted is

         N_o = 0.8754 * 12.5

         N_o = 10.94 \ moles

The volume is

      V_o  = \frac{10.94 * 0.08206 *623}{0.967}

      V_o  = 579 \ L

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

         V_e = V_o * \frac{0.79}{0.21}

Substituting values

       V_e = 579 * \frac{0.79}{0.21}

       V_e = 2178 \ L

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Limitation for osmosis experiment
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How many visible impact craters does the moon contain?
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I believe the answer is B. a million or more.
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3 years ago
Calculate the number of atoms in 35 g of carbon. Calculate the number of atoms in 75 g of bromine.
nata0808 [166]

Answer:

Explanation:

To calculate the number of atoms in a formula, the weight of a sample, its atomic mass from the periodic table and a constant known as Avogadro’s number are needed.

Step 1: Find the Molar Mass of the Formula

Find a periodic table of elements to find the molar mass of your sample. If your sample is made of one element, like copper, locate the atomic mass on the periodic table. Atomic mass is usually listed below the symbol for that element.

The atomic mass of carbon and bromine is 63.55 atomic mass units. This formula mass is numerically equal to the molar mass in grams/mole, and this means copper is 63.55 grams/mole.

Whether you use an individual element like copper or a molecule, the procedure for finding the atoms in a formula remains the same.

Find the molar mass of carbon and bromine on the periodic table: 63.55 grams/mole.

Understanding the Mole: The mole (often abbreviated as mol) listed above is a unit of measurement. If you sold eggs, you would talk about them in the dozens, not one by one.

A mole is a certain amount, too. If chemists want to speak about incredibly small atoms and molecules, an amount far greater than a dozen is needed. A mole is Avogadro’s number of items: 6.022 × 1023.

1 mole of C atoms = 6.022 × 1023 C atoms

1 mole of Br atoms = 6.022 × 1023 S atoms

1 mole of CBrmolecules = 6.022 × 1023 CBr molecules

1 mole of pennies = 6.022 × 1023 pennies

To give an idea of how large this number is, 1 mole of pennies would be enough money to pay all the expenses of each country on earth for about the next billion years.

Step 2: Find the Number of Moles

The example is 35 grams of C and . Change that into moles using the molar mass you found in Step 1. Chemists use ratios for this calculation.

Start with what you know and add in the molar mass ratio, so the units will cancel:

35g of C× 1 mol C/ 12 g C = 2.92mol of C

Step 3: Convert Moles to Atoms Using Avogadro’s Number

Once the amount of moles is known, the number of atoms in the formula may be calculated using Avogadro’s number. Again, use the ratio format.

Notice the number of moles is used from Step 2 to start the calculation from moles to atoms:

2.92 mol of C× 6.022 x 1023 atoms / 1 mol of C = 3.13 x 1023 molecules

To answer your example question, there are 3.13 × 1023 atoms in 32.80 grams of carbon.

Steps 2 and 3 can be combined. Set it up like the following:

32.80 g of C × 1 mol Cu / 159.17 g C × 6.022 x 1023 atoms / 1 mol of C = 3.13 x 1023 atoms in 32.80 grams of carbon

Several online sites have a number of atoms calculator. One is the Omni Calculator and is listed in the Resources section, but you’ll still need to know how to calculate molar mass (Step 1).

Historical Note: Why Is It Called Avogadro's Number?

Avogadro’s number is named after Amedeo Avogadro (1776-1856), an Italian scientist that hypothesized that equal volumes of gases at the same temperature and pressure will have the same number of particles.

Avogadro did not propose the constant, 6.022 ×1023, but because of his contributions to science, the constant was named after him. Incidentally, Avogadro first introduced his gas theory in 1811, and it was ignored for 50 years.

5 0
3 years ago
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