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dem82 [27]
2 years ago
15

12.3 moles of sodium is what mass of Na?​

Chemistry
1 answer:
olasank [31]2 years ago
8 0

Answer:

1 mole of Na = mass of 22.99 g, 1 mole of Si = mass of 28.09 g.

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How many liters of chlorine gas can react with 56.0 grams of calcium metal at standard temperature and pressure? Show all of the
Doss [256]
The balanced chemical reaction is:

<span>Ca + Cl2 =  CaCl2
</span>
We are given the amount of calcium metal to be used for this reaction. This will be the starting point for the calculations.

56 g Ca ( 1 mol Ca / 40.08 g Ca) (1 mol Cl2 / 1 mol Ca) ( 22.414 L Cl2 / 1 mol Cl2 ) = 31.32 L Cl2 gas produced from the reaction
5 0
3 years ago
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I need help solving this for chemistry. Don’t know where to start:/
Juli2301 [7.4K]
1 electron has charge =1.602* 10⁻¹⁹ C
1 mole of electrons have 1.602* 10⁻¹⁹*6.02*10²³C = 9.64*10⁴ C/1mol

One ion Co²⁺   takes 2e⁻ to become Co⁰.
1 mol of Co²⁺  ions take 2 mole of e⁻ to become Co⁰, so
 0.30 mol Co²⁺  ions take mole of 0.60 mol e⁻ to become Co⁰

9.64*10⁴(C/1mol) *0.60 (mol)≈ 5.8 *10⁴ Coulombs.
Correct answer is C
8 0
3 years ago
Analysis of a gaseous chlorofluorocarbon, CClxFy, shows that it contains 11.79% C and 69.57% Cl. In another experiment, you find
uranmaximum [27]

Answer:

The molecular formula = C_2Cl_{4}F_2

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

% of C = 11.79

Molar mass of C = 12.0107 g/mol

<u>% moles of C = 11.79 / 12.0107 = 0.9816</u>

% of Cl = 69.57

Molar mass of Cl = 35.453 g/mol

<u>% moles of Cl = 69.57 / 35.453 = 1.9623</u>

Given that the gaseous chlorofluorocarbon only contains chlorine, flourine and carbon. So,

% of F = 100% - % of C - % of C = 100 - 11.79 - 69.57 = 18.64

Molar mass of F = 18.998 g/mol

<u>% moles of F = 18.64 / 18.998 = 0.9812</u>

Taking the simplest ratio for C, Cl and F as:

0.9816 : 1.9623 : 0.9812

= 1 : 2 : 1

The empirical formula is = CCl_2F

Also, Given that:

Pressure = 21.3 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 21.3 / 760 = 0.02803 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

Volume = 458 mL  = 0.458 L (1 mL = 0.001 L)

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.02803 atm × 0.458 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 0.00052445 moles

Given that :  

Amount  = 0.107 g  

Molar mass = ?

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.00052445= \frac{0.107\ g}{Molar\ mass}

Molar\ mass= 204.0233\ g/mol

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12.0107 + 2×35.453 + 1×18.998 = 101.9147 g/mol

Molar mass = 204.0233 g/mol

So,  

Molecular mass = n × Empirical mass

204.0233 = n × 101.9147

⇒ n = 2

<u>The molecular formula = C_2Cl_{4}F_2</u>

6 0
3 years ago
Given the reaction that occurs in an electrochemical cell:
KatRina [158]

Answer:

c) +2 to 0

Explanation:

SO4 has a charge of -2, so the Cu attached to that has to be a +2 since the polyatomic molecule has no overall charge

Cu(s) is a solid metal and they have no charge, therefore it is zero

Copper undergoes Oxidation (gain of electrons)

8 0
3 years ago
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Determine the mass of sodium carbonate required to produce 23.4g of sodium chloride when it reacts with excess hydrochloric acid
Leni [432]
The balanced chemical reaction for the described reaction above is,
                          Na2CO3 + 2HCl ---> 2NaCl + H2CO3
From the reaction, 1 mole of Na2CO3 is needed to produce 2 moles of NaCl. In terms of mass, 106 grams of Na2CO3 are needed to produce 116.9 grams of NaCl. From this,
                (23.4 g NaCl) x (106 g Na2CO3 / 116.9 NaCl = 21.22 g Na2CO3
Thus, approximately 21.22 g Na2CO3 is needed for the desired reaction. 
8 0
3 years ago
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