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Morgarella [4.7K]
2 years ago
7

What else is produced during the replacement reaction of silver nitrate and potassium sulfate? 2AgNO3 + K2SO4 Ag2SO4 + ________

Chemistry
2 answers:
NemiM [27]2 years ago
7 0

Answer:

c

Explanation:

OleMash [197]2 years ago
5 0

Answer:

c

Explanation:

You might be interested in
A 44.8 g rhodium sample contains how many rhodium atoms
Tresset [83]

Answer:

2.63*10^23

Explanation:

1 mol rhodium = 102.91

44.8g/1 mol * 1 mol/ 102.91mol * 6.022*10^23/1 mol =

2.63*10^23

6 0
3 years ago
Write the balanced molecular chemical equation for the reaction
Ierofanga [76]

The balanced molecular chemical equation for the reaction will be expressed as Cs₂CO₃ + Mg(NO₃)₂ -> 2CsNO₃ + MgCO₃

  • For any chemical equation to be balanced, the number of moles of elements in the reactants must be equal to that of the product.

  • According to the question, we are to write a balanced equation for the reaction  in aqueous solution for cesium carbonate and magnesium  nitrate
  • The chemical formula for Cesium carbonate is Cs₂CO₃
  • The chemical formula for magnesium  nitrate is Mg(NO₃)₂

Hence the balanced molecular chemical equation for the reaction will be expressed as Cs₂CO₃ + Mg(NO₃)₂ -> 2CsNO₃ + MgCO₃

Learn more here: brainly.com/question/11904811

4 0
3 years ago
Read 2 more answers
Kepler’s investigations of the movement of the planets explained which of the following?
morpeh [17]

Answer:

The earth‘s orbit

Explanation:

8 0
3 years ago
Calculate the volume of 1M NaOH required to neutralize 200 cc of 2M HCI. What mass of sodium
Nezavi [6.7K]

NaOH+HCl-> NaCl+H2O

1 mole of NaOH

1 mole of HCl.

To calculate volume of NaOH

CaVa/CbVb= Na/Nb

Where Ca=2M

Cb=1M

Va=200cm³

Vb=xcm³

Substitute into the equation.

2×200/1×Vb=1/1

400/Vb=1/1

Cross multiply

Vb×1=400×1

Vb=400cm³

To calculate the mass of sodium chloride, NaCl from the neutralization rxn.

Mole of NaCl=1

Molar mass of NaCl= 23+35.5=58.5

Mass=xgrammes.

Mass of NaCl=Number of moles × Molar mass.

Substitute

Mass of NaCl= 1×58.5

=58.5g

This is what I could come up with.

5 0
3 years ago
What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2
atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}

Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

Mole of AlCl_3 = Mole of Al * Mole ratio

Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = 133.34 \frac{g}{mol}

Mass of AlCl3 can be calculated using mole formula as:

1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}

Multiplying both side by 133.34 \frac{g}{mol}

1.32 mole  * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34  \frac{g}{mol}

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

Mole of AlCl_3 = mole of Cl_2 * mole ratio

Mole of AlCl_3 = 0.55  mole  * \frac{2}{3}

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}

Multiplying both side by

133.34 \frac{g}{mol}

0.367 mol of AlCl_3 * 133.34 \frac{g}{mol}  = \frac{mass(g)}{133.34\frac{g}{mol}}   * 133.34 \frac{g}{mol}

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

7 0
3 years ago
Read 2 more answers
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