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OLga [1]
3 years ago
15

Classify the chemical reaction shown here: CrF3+H3PO4→CrPO4+3HF

Chemistry
1 answer:
Lana71 [14]3 years ago
5 0
The answer 
<span>he chemical reaction CrF3+H3PO4 → CrPO4 + 3HF
the main formula is 
when we observe a reaction as follow:
XY  +  ZT --------------- XT + YZ  
this is classified as double replacement reactions, this means: </span>two ionic compounds exchange ions, producing 2 new ionic compounds.
in our case we have CrF3+H3PO4 → CrPO4 + 3HF
so the classification is double-replacement reactions
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CO2 + H2O - H2CO3<br> The reaction above is classified as
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Answer:

Nonmetal oxides react with water to form oxyacids. Ex. CO2 + H2O → H2CO3 Page 3 Decomposition - compound (reactant) breaks down into 2 or more simpler substances.

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3 years ago
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3.Write the chemical equation for the reaction when methane burns in (2)
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Answer:

a)CH4 + O2 = CO2 + H2O

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3 years ago
Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2NH3(g) 3N2O(g)4N2(g) 3H2O
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Answer: \Delta H^{0} = -879.15 kJ/mol

Explanation: <u>Heats</u> <u>of</u> <u>formation</u> is the amount of heat necessary to create 1 mol of a compound from its molecular constituents. The basic conditions the substance is formed is at standard conditions: 1 atm and 25°C. Each compound has its own heat of formation per mol of compound (kJ/mol), but to an element is assigned a value of zero.

<u>Standard</u> <u>Enthalpy</u> <u>Change</u> is defined as the heat absorbed or released when a reaction takes place. It can be positive or negative, which means reaction is endothermic or exothermic, respectively.

Enthalpy change is calculated as the difference between the sum of heat formation of products and the sum of heat formation of the reactants:

\Delta H^{0}=\Sigma H^{0}_{f}_{(products)}-\Sigma H^{0}_{f}_{(reactants)}

For the reaction

   2NH₃   +      3N₂O   →      4N₂    +     3H₂O

2(-46.2)   +     3(82.05)      4(0)    +    3(-241.8)

\Delta H^{0}=3(-241.8)-[ 2(-46.2)+3(82.05)]

\Delta H^{0}=-725.4-153.75

\Delta H^{0}=-879.15

<u>The standard enthalpy change for the reaction is </u>\Delta H^{0}=-879.15<u> kJ</u>

8 0
2 years ago
which example would most likely decrease friction? using cleats, rather than gym shoes, on a field using shaving cream, rather t
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4 0
3 years ago
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Determine the molarity for each of the following solution solutions:
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Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of H_2SO_4 solution is, 0.00525 mole/L

(c)The molarity of Al(NO_3)_3 solution is, 0.0612 mole/L

(d)The molarity of CuSO_4.5H_2O solution is, 7.61 mole/L

(e)The molarity of Br_2 solution is, 0.0565 mole/L

(f)The molarity of C_2H_5NO_2 solution is, 0.0113 mole/L

Explanation :

<u>(a) 1.457 mol of KCl in 1.500 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Solute is KCl.

\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L

The molarity of KCl solution is, 0.9713 mole/L

<u>(b) 0.515 gram of H_2SO_4, in 1.00 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Solute is H_2SO_4

Molar mass of H_2SO_4 = 98 g/mole

\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L

The molarity of H_2SO_4 solution is, 0.00525 mole/L

<u>(c) 20.54 g of Al(NO_3)_3 in 1575 mL of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Solute is Al(NO_3)_3

Molar mass of Al(NO_3)_3 = 213 g/mole

\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L

The molarity of Al(NO_3)_3 solution is, 0.0612 mole/L

<u>(d) 2.76 kg of CuSO_4.5H_2O in 1.45 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Solute is CuSO_4.5H_2O

Molar mass of CuSO_4.5H_2O = 250 g/mole

\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L

The molarity of CuSO_4.5H_2O solution is, 7.61 mole/L

<u>(e) 0.005653 mol of Br_2 in 10.00 ml of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

Solute is Br_2.

\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L

The molarity of Br_2 solution is, 0.0565 mole/L

<u>(f) 0.000889 g of glycine, C_2H_5NO_2, in 1.05 mL of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Solute is C_2H_5NO_2

Molar mass of C_2H_5NO_2 = 75 g/mole

\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L

The molarity of C_2H_5NO_2 solution is, 0.0113 mole/L

5 0
3 years ago
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