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andrezito [222]
2 years ago
5

can someone please help me with this question for chemistry. What is the number of moles in 1216 g Sr3(PO4)2? the 3,4, and 2 and

small numbers that go under the letters btw
Chemistry
1 answer:
Alisiya [41]2 years ago
8 0
2.68 mol should be the answer
You might be interested in
How many moles of glucose does 1.2 x 10^24 molecules represent
sammy [17]
<h3>Answer:</h3>

2.0 mol C₆H₁₂O₆

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

1.2 × 10²⁴ molecules C₆H₁₂O₆ (glucose)

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 1.2 \cdot 10^{24} \ molecules \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{6.022 \cdot 10^{23} \ molecules \ C_6H_{12}O_6})
  2. Divide:                                                                                                                      \displaystyle 1.99269 \ mol \ C_6H_{12}O_6

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.99269 mol C₆H₁₂O₆ ≈ 2.0 mol C₆H₁₂O₆

3 0
3 years ago
For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia
Fittoniya [83]

Answer : The value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o_{(Cu^{2+}/Cu)} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}

Putting values in above equation, we get:

1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)

E^o_{(Cu^{2+}/Cu)}=0.34V

Hence, the value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

8 0
2 years ago
2. Fe(s) + O2(g) → Fe3O4(s) a. When 13.54 g of O2 is mixed with 12.21 g of Fe, which is the limiting reactant? b. What mass in g
julsineya [31]

The percent yield shows the extent to which the reactants are converetd into products . The limiting reactant is used up in the reaction.

<h3>What is a limiting reactant?</h3>

A limiting reactant is the reactant that is in the least amount in the system.  Now;

Number of moles of Fe =  12.21 g/56 g/mol = 0.22 moles

Number of moles of O2 = 13.54 g/32 g/mol = 0.42 moles

Balanced reaction equation;

3Fe(s) + 2 O2(g) = Fe3O4(s)

If 3 moles of Fe reacts with 2 mole of O2

0.22 moles of Fe reacts with  0.22 moles * 2 mole/3 moles = 0.15 moles

Hence, Fe is the limiting reactant

If 3 mole of Fe produces 1 mole of  Fe3O4(s)

 0.22 moles  of O2   produces   0.22 moles * 1 mole/3 moles of  Fe3O4(s) = 0.1073 moles

Mass of  Fe3O4(s) =0.1073 moles  * 232 g/mol =16.9 g

Number of moles of excess reactant = 0.42 moles - 0.15 moles = 0.27 moles

Mass of excess reactant = 0.27 moles * 32 g/mol = 8.64 g

percent yield = 15.88 g/16.9 g * 100/1

= 93.4%

Learn more about percent yiled: brainly.com/question/13463225

8 0
2 years ago
Calculate the rate constant at 200.°C for a reaction that has a rate constant of 8.30 × 10−4 s−1 at 90.°C and an activation ener
Sergeu [11.5K]

Answer:

23.0 s⁻¹ is rate constant

Explanation:

Using the Arrhenius equation:

k = A * e^(-Ea/RT)

Where k is rate constant

A is frequency factor (1.5x10¹¹s⁻¹)

Ea is activation energy = 55800J/mol

R is gas constant (8.314J/molK)

And T is absolute temperature (24°C + 273 = 297K)

Replacing:

k = 1.5x10¹¹s⁻¹ * e^(-55800J/mol/8.314J/molK*297K)

k = 1.5x10¹¹s⁻¹ * 1.53x10⁻¹⁰

k = 23.0 s⁻¹ is rate constant    i hope this helpsss

Explanation:

4 0
3 years ago
Which subatomic particle is not found at the center of the atom?
Snezhnost [94]
The answer is nucleus
8 0
3 years ago
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