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3 years ago

Na-23 has 12 neutrons.what is its atomic number? A.11 B.12 C.23 D.34

1 answer:

4 0

Google says 11 lol full google answer:

Sodium-23 atom is the stable isotope of sodium with relative atomic mass 22.989770, 100 atom percent natural abundance and nuclear spin 3/2. A member of the alkali group of metals. It has the atomic symbol Na, atomic number 11, and atomic weight 23.

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Consider the following reaction.

Ivan

Answer:

162 g Fe₂O₃

Explanation:

To find the mass of Fe₂O₃, you need to (1) convert grams C to moles C (via molar mass from periodic table), then (2) convert moles C to moles Fe₂O₃ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles Fe₂O₃ to grams (via molar mass). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to reflect the given value.

Molar Mass (C): 12.011 g/mol

2 Fe₂O₃(s) + 3 C(s) ---> 4 Fe(s) + 3 CO₂(g)

Molar Mass (Fe₂O₃): 2(55.845 g/mol) + 3(15.998 g/mol)

Molar Mass (Fe₂O₃): 159.684 g/mol

18.3 g C 1 mole 2 moles Fe₂O₃ 159.684 g
-------------- x ---------------- x ------------------------- x ----------------- = 162 g Fe₂O₃
12.011 g 3 moles C 1 mole

5 0

2 years ago

3. A measurement of the freezing temperature of a solution allowsyou to calculate the concentration of the solution. What else d

Harman [31]

Answer:

Osmotic pressure and boiling point elevation

Explanation:

In the the osmotic pressure one can determine the molar mass of a solid by calculating the number of moles from the Morality formula which involves the volume of the solution.

In the boiling point elevation you can determine the number of moles of the solute in the solution by using the Molality formula.

5 0

3 years ago

It takes 60 mL of 0.20 M of sodium hydroxide (NaOH) to neutralize 25 mL of carbonic acid (H2CO3) for the following chemical reac

irina1246 [14]

If It takes 60mL of 0.20M of sodium hydroxide (NaOH) to neutralize 25 mL of carbonic acid ( $H_2CO_3$ ), then the concentration of the carbonic acid is 0.24M

The reaction between NaOH solution and $H_2CO_3$ is written below

$2NaOH + H_2CO_3 \rightarrow Na_2CO_3 + 2H_2O$

Volume of NaOH, $V_B$ = 60 ml

Volume of $H_2CO_3$ , $V_A=25 ml$

Molarity of $H_2CO_3$ , $C_A=?$

Molarity of NaOH, $C_B=0.20M$

Number of moles of $H_2CO_3$ , $n_A=1$

Number of moles of NaOH, $n_B=2$

The mathematical equation for neutralization reaction is:

$\frac{C_AV_A}{C_BV_B} =\frac{n_A}{n_B}$

Substitute $C_B=0.2 M$ , $n_A=1$ , $n_B=2$ , $V_B$ = 60ml, and $V_A=25 ml$ into the equation above in order to solve for $C_A$

$\frac{C_A \times 25}{0.2 \times 60}=\frac{1}{2} \\\\50C_A=12\\\\C_B=\frac{12}{50} \\\\C_B=0.24M$

Therefore, the concentration of the carbonic acid is 0.24M

Learn more here: brainly.com/question/25943090

3 0

2 years ago

A sample of Manganese (II) chloride has a mass of 19.8 grams before heating, and 12.6 grams after heating until all the water is

IRINA_888 [86]

Answer:

no. of water molecules associated to each molecule of $MnCl_2$ = 4

Explanation:

Mass of $MnCl_2$ before heating = 19.8 g

Mass of $MnCl_2$ after heating = 12.6 g

Difference in mass of $MnCl_2$ before and after heating

= 19.8 - 12.6 = 7.2 g

Difference in mass corresponds to mass of water driven out.

Molar mass of water = 18 g/mol

No. of moles of water = $\frac{7.2}{18} = 0.4\ mol$

Mass of $MnCl_2$ obtained after heating is mass of anhydrous $MnCl_2$ .

Mass of anhydrous $MnCl_2$ = 12.6 g

Molar mass of $MnCl_2$ = 125.9 g/mol

No. of mol of anhydrous $MnCl_2$ = $\frac{125.9}{125.9} = 0.1\ mol$

so,

0.1 mol of $MnCl_2$ have 0.4 mol of water

1 mol of $MnCl_2$ will have = $\frac{0.4}{0.1} =4\ mol$

Hence, no. of water molecules associated to each molecule of $MnCl_2$ = 4

5 0

3 years ago

An unknown amount of Al203 decomposed producing 215 g of solid aluminum. 2Al2O3=4Al+3O2 How many grams of oxygen gas should be p

natulia [17]

Answer:

191.11 grams of oxygen gas should be produced.

Explanation:

The balanced reaction is:

2 Al₂O₃ → 4 Al + 3 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Al₂O₃: 2 moles
Al: 4 moles
O₂: 3 moles

Being the molar mass of each compound:

Al₂O₃: 102 g/mole
Al: 27 g/mole
O₂: 32 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Al₂O₃: 2 moles* 102 g/mole= 204 grams
Al: 4 moles* 27 g/mole= 108 grams
O₂: 3 moles* 32 g/mole= 96 grams

Then you can apply the following rule of three: if by stoichiometry 108 grams of aluminum are produced along with 96 grams of oxygen, 215 grams of aluminum are produced along with how much mass of oxygen?

$mass of oxygen=\frac{215 grams of aluminum*96 grams of oxygen}{108grams of aluminum}$

mass of oxygen= 191.11 grams

<u><em>191.11 grams of oxygen gas should be produced.</em></u>

6 0

2 years ago