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Reika [66]
3 years ago
7

What are the coefficients that would correctly balance this equation? __Zn+__K2CrO4 > __K + __ZnCrO4

Chemistry
1 answer:
saul85 [17]3 years ago
3 0

D. 1,1,2,1

<h3>Further explanation  </h3>

Equalization of chemical reactions can be done using variables. Steps in equalizing the reaction equation:  

  • 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c, etc.  
  • 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index (subscript) between reactant and product  
  • 3. Select the coefficient of the substance with the most complex chemical formula equal to 1  

Reaction(unbalanced)

Zn+K₂CrO₄ ⇒ K + ZnCrO₄

Give coefficient

aZn+K₂CrO₄ ⇒ bK + cZnCrO₄

Zn, left=a, right=c⇒a=c

K, left=2, right=b⇒b=2

Cr, left=1, right=c⇒c=1⇒a=1

O,left=4,right=4c⇒4c=4⇒c=1

Reaction(balanced0

Zn+K₂CrO₄ ⇒ 2K + ZnCrO₄

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How many grams of nacl are in 250 g of a 15.0% (by weight) solution?
Gelneren [198K]

Answer: -

37.5 g of NaCl are in 250 g of a 15.0% (by weight) solution.

Explanation: -

Total mass = 250 g

Percentage of NaCl = 15.0 %

= \frac{Number of grams of NaCl present}{total mass}

Thus \frac{15}{100} = \frac{Number of grams of NaCl present}{250g}

Number of grams of NaCl present = \frac{15 x 250g}{100}

= 37.5 g

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4 years ago
Is this sentence true or false ? hydrogen is considered to be a metal
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For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car
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Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

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

  • Na₂CO₃: 1 mole
  • Ca(NO₃)₂: 1 mole
  • CaCO₃: 1 mole
  • NaNO₃: 2 mole

Being the molar mass of the compounds:

  • Na₂CO₃: 106 g/mole
  • Ca(NO₃)₂: 164 g/mole  
  • CaCO₃: 100 g/mole
  • NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

  • Na₂CO₃: 1 mole* 106 g/mole= 106 g
  • Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
  • CaCO₃: 1 mole* 100 g/mole= 100 g
  • NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of  CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of  CaCO₃?

mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}

mass of CaCO₃= 74.81 grams

<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>

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4 years ago
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