How much iron is present in 7.59 g of iron(iii) oxide? answer in units of g?

1 answer:

4 0

You need the unit of g however you must convert to moles before you can use the mole ratio to find the moles of iron, you used the molar mass of iron to find the grams of iron. since F e 2 o 3

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Given that the molar mass of NaCl is 58.44 g/mol, what is the molarity of a solution that contains 87.75 g of NaCl in 500. mL of

Stells [14]

Given information:

Mass of NaCl (m) = 87.75 g

Volume of solution (V) = 500 ml = 0.5 L

Molar mass of NaCl (M) = 58.44 g/mol

To determine:

The molarity of NaCl solution

Explanation:

Molarity is defined as the number of moles of solute(n) dissolved per liter of solution (V)

i.e. M = moles of solute/liters of solution = n/V

Moles of solute (n) = mass of solute (m)/molar mass (M)

moles of NaCl = 87.75 g/58.55 g.mol-1 = 1.499 moles

Therefore,

Molarity of NaCl = 1.499 moles/0.5 L = 2.998 moles/lit ≅ 3 M

Ans: (D)

4 0

3 years ago

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The half-life of na-24 is 15 hours . when there are 1000 atoms of na-24 in a sample , a scientist starts a stopwatch . the scien

marin [14]

125 Each half life it divides by 2 the amount
1000/2=500
500/2=250
250/2=125

8 0

3 years ago

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Just as heat can denature enzymes, so can a change in the pH. Enzymes that normally work in an acidic environment in the body ca

nikitadnepr [17]

Answer:

It will create an alkaline environment and can lead to denaturation of acidic enzymes.

Explanation:

Taking too much antacids will significantly increase the pH of the stomach and create an alkaline environment within it. Consequently, enzymes that normally work in acidic environments in the body can become denatured or rendered inactive. This will create another problem for the individual.

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In a hydrogen atom, the electron and the proton are seperated by about 0.5 angstroms, 5.0×10−11 m. what is the acceleration of a

nadya68 [22]

The charge of proton and electron is equal in magnitude that is $1.6\times 10^{-19} C$ . The distance between electron and proton is $0.5 \AA$ or $5\times 10^{-11} m$ .

Force exerted on electron by proton is calculated as follows:

$F= \left | \frac{kq_{e}q_{p}}{r^{2}} \right |$

Here, k is Coulomb's constant, $q_{p}$ is charge on proton, $q_{e}$ charge on electron and r is distance between them.

Putting the values,

$F= \left | \frac{(8.99\times 10^{9}N m^{2}/C^{2})(1.6\times 10^{-19} C)(1.6\times 10^{-19} C)}{(5\times 10^{-11} m)^{2}} \right |=9.2\times 10^{-8} N$