1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
garik1379 [7]
2 years ago
5

How many formula units make up 24.2 g of magnesium chloride (MgCl2)? Help!!

Chemistry
1 answer:
NNADVOKAT [17]2 years ago
6 0

Answer:

Approximately 1.53 \times 10^{23} formula units (0.254\; \rm mol).

Explanation:

Refer to a modern periodic table for the relative atomic mass of magnesium (\rm Mg) and chlorine (\rm Cl):

  • \rm Mg: 24.305.
  • \rm Cl: 35.45.

In other words, the mass of 1\; \rm mol of \rm Mg atoms would be (approximately) 24.305\; \rm g.

Likewise, the mass of 1\; \rm mol of \rm Cl atoms would be approximately 35.45\; \rm g.

One formula unit of the ionic compound \rm MgCl_{2} includes exactly as many atoms as there are in the given formula. The formula mass of a compound is the mass of 1\; \rm mol of the formula units of this compound.

The formula \rm MgCl_{2} includes one \rm Mg atom and two \rm Cl atoms.

Hence, every formula unit of \rm MgCl_{2} \! would include the same number of atoms: one \rm Mg\! atom and two \rm Cl\! atoms. There would be 1\; \rm mol of \rm Mg atoms and 2\; \rm mol of \rm Cl atoms in 1\; \rm mol\! of \rm MgCl_{2} formula units.

Thus, the mass of 1\; \rm mol\! of \rm MgCl_{2} formula units would be equal to the mass of 1\; \rm mol of \rm Mg atoms plus the mass of 2\; \rm mol of \rm Cl atoms. (The mass of 1\; \rm mol\!\! of each atom could be found from the relative atomic mass of each element.)

\begin{aligned}& M({\rm MgCl_{2}}) \\ =\; & 24.305\; {\rm g \cdot mol^{-1}} + 2\times {\rm 35.45 \; \rm g \cdot mol^{-1}} \\ =\; & 95.205\; \rm g \cdot mol^{-1}\end{aligned}.

In other words, the formula mass of \rm MgCl_{2} is 95.205\; \rm g \cdot mol^{-1}.

Therefore, the number of formula units in m = 24.2\; \rm g of \rm MgCl_{2} would be:

\begin{aligned}n &= \frac{m({\rm MgCl_{2}})}{M({\rm MgCl_{2}})} \\ &= \frac{24.2\; \rm g}{95.205\; \rm g\cdot mol^{-1}} \\ & \approx 0.254\; \rm mol\end{aligned}.

Multiple n by Avogadro's Number N_{A} \approx 6.022 \times 10^{23}\; \rm mol^{-1} to estimate the number of formula units in 0.254\; \rm mol:

\begin{aligned}N &= n \cdot N_{A} \\ &\approx 0.254\; \rm mol \times 6.022 \times 10^{23}\; \rm mol^{-1} \\ &\approx 1.53\times 10^{23}\end{aligned}.

You might be interested in
Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result
Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(c) 4.886

(d) 5.363

(e) 5.570

(f)  12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only  the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

(c) = mol HA reacted = 0.100 L x 0.14 mol/L = 0.014 mol

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) =  4.886

(d) mol HA reacted = 150 x 10⁻³ L  x  x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) =  5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH  = √(kb x (A⁻)  where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA,  200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒   Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

4 0
2 years ago
HELP ME THIS IS DUE TODAY PLEAS HELP SO I DONT FAIL
bija089 [108]
For the first one it’s 69 just count the little lines from the side.
3 0
3 years ago
2. Heat is:
nlexa [21]
A. Thermal energy good job
5 0
3 years ago
Read 2 more answers
Methane (CH4), ammonia (NH3), and oxygen (O2) can react to form hydrogen cyanide (HCN) and water according to this equation:
egoroff_w [7]

The grams  of  oxygen that are required  to produce 1  mole  of H₂O  is 16 g ( answer  B)

<u><em> calculation</em></u>

2 CH₄  + 2NH₃ +3 O₂ → 2HCN  + 6H₂O

step 1: use the mole ratio to find moles of O₂

from equation above the  mole ratio  of O₂: H₂O  is 3:6 therefore the moles of O₂  = 1 mole x3/6 =0.5 moles

step  2: find   mass of O₂

mass= moles x molar mass

from periodic table the molar mass  of O₂ = 16 x2= 32 g/mol

mass O₂ = 0.5 moles x 32 g/mol = 16 g (answer B)

6 0
3 years ago
Read 2 more answers
How many atoms are in the 1.5 moles of tin atoms?
Anastasy [175]

Answer:

option C = 9.0 x 10²³ atoms

Explanation:

Data Given:

no. of moles of tin (Sn) atoms = 1.5 moles

no. of tin (Sn) atoms = ?

Solution:

Formula used to find number of atoms

                  no. of moles = no. of atoms / Avogadro's number

Rearrange the above equation:

          no. of atoms =   no. of moles x Avogadro's number . . . . . . (1)

Where

Avogadro's number = 6.022 x 10²³

Put values in equation 1

           no. of atoms = 1.5 x 6.022 x 10²³

           no. of atoms = 9.033 x 10²³

Round the figure = 9.0 x 10²³ atoms

So option C is correct

6 0
3 years ago
Other questions:
  • Define the term compound
    12·1 answer
  • A substance that enters into a chemical reaction is called a(n) _____.
    7·2 answers
  • Which atomic model was proposed as a result of J. J. Thomson’s work?
    7·1 answer
  • Which ions have ten electrons in the outermost subshell?
    13·1 answer
  • Explain chromatography​
    15·1 answer
  • How many joules are required to change 40 grams of water to steam at the boiling point?
    10·1 answer
  • Which two of these are also pure substances? Silicon <br>Brass <br>Gold <br>Steel <br>Rubber​
    12·2 answers
  • the wind pushes a paper cup along the sand at a beach. the cup has a mass of 0.025 kg and accelerates at a rate of 5 m/s2. how m
    5·1 answer
  • If you make a solution using 1.0 mol KBr in 1.0 kg water, how will the vapor pressure of this solution compare to the vapor pres
    9·2 answers
  • How many ml of naoh was used to reach the equivalence point?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!