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vaieri [72.5K]
3 years ago
11

Which of the following reactions below could be properly categorized as a synthesis reaction?

Chemistry
1 answer:
kari74 [83]3 years ago
3 0

Answer:

A synthesis reaction is a type of reaction in which multiple reactants combine to form a single product. Synthesis reactions release energy in the form of heat and light, so they are exothermic. An example of a synthesis reaction is the formation of water from hydrogen and oxygen.

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In the example in the lesson, 0.10 mole of sodium chloride or magnesium chloride or aluminum chloride was added to one liter of
Elza [17]

Answer:

1x10⁻⁴ moles of Cl⁻ in NaCl solution

2x10⁻⁴ moles of Cl⁻ in MgCl₂ solution

3x10⁻⁴ moles of Cl⁻ in AlCl₃ solution

Explanation:

Let's produce the reactions of each salt

NaCl  → Na⁺ +  Cl⁻

0.1 m                 0.1m

MgCl₂ → Mg²⁺  +  2Cl⁻

0.1m                      0.2m

AlCl₃ → Al³⁺  +  3Cl⁻

0.1 m                 0.3m

0.1, 0.2 and 0.3 are the moles of each chloride in each solution and this moles are added to one liter of solvent.

1 L = 1000mL

Let's prepare the rule of three for each

1000 mL ____ 0.1 m ____ 0.2 m _____ 0.3m

1 mL______ 1x10⁻⁴ m ___ 2x10⁻⁴ m _____ 3x10⁻⁴ m

4 0
3 years ago
Describe 2 physical changes that could occur with a piece of paper
n200080 [17]

Answer:

Physical=Ripping

Chemical=Burning

Explanation:

I may be wrong. But I would say these are the answers.

3 0
3 years ago
Read 2 more answers
The enthalpy change for converting 1.00 mol of ice at -25.0 ∘c to water at 90.0∘c is ________ kj. the specific heats of ice, wat
liubo4ka [24]

Answer : The enthalpy change for converting 1 mole of ice at -25.0^oC to water at 90^oC is, 7.712 KJ

Solution :

Process involved in the calculation of enthalpy change :

(1):ice(-25^oC)\rightarrow ice(0^oC)\\\\(2):ice(0^oC)\rightarrow water(0^oC)\\\\(3):water(0^oC)\rightarrow water(90^oC)

Now we have to calculate the enthalpy change.

\Delta H=[m\times c_{ice}\times (T_2-T_1)]+\Delta H_{fusion}+[m\times c_{water}\times (T_3-T_2)]

where,

\Delta H = enthalpy change

m = mass of water = 1mole\times 18g/mole=18g

c_{ice} = specific heat of ice = 2.09 J/gk

c_{water} = specific heat of water = 4.18 J/gk

\Delta H_{fusion} = enthalpy change for fusion = 6.01 KJ/mole = 0.00601 J/mole

conversion : 0^oC=273k

T_1 = initial temperature of ice = 0^oC=273k

T_2 = final temperature of ice = -25^oC=273+(-25)=248k

T_3 = initial temperature of water = 0^oC=273k

T_4 = final temperature of water = 90^oC=273+90=363k

Now put all the given values in the above expression, we get

\Delta H=[18g\times 2.09J/gK\times (273-248)k]+0.00601J+[18g\times 4.18J/gK\times (363-273)k]

\Delta H=7712.106J=7.712KJ     (1 KJ = 1000 J)

Therefore, the enthalpy change for converting 1 mole of ice at -25.0^oC to water at 90^oC is, 7.712 KJ

3 0
3 years ago
Write the IUPAC name for the compound that reacts with vegetable oil to produce biodiesel
sleet_krkn [62]
Methanol  or  <span>methyl alcohol</span>
6 0
3 years ago
What is the mass of a sample of iron that has had 300.0 j applied to it and heats up from 20.0 degrees Celsius to 40.0 degrees C
yuradex [85]

Answer:

The mass of a sample of iron that has had 300 J applied to it and heats up from 20 degrees Celsius to 40 degrees Celsius is 32.61 grams.

Explanation:

Calorimetry is the measurement and calculation of the measurement of heat changes exchanged by a body or a system produced in physical and chemical processes.

The sensible heat of a body is the amount of heat received or transferred by a body to produce a change in temperature but without a change in physical state.

The sensible heat in a constant pressure is calculated by:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c, and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial)

In this case:

  • Q= 300 J
  • c= 0.46 \frac{J}{g*C}
  • m= ?
  • Tfinal= 40 C
  • Tinitial= 20 C

Replacing:

300 J= 0.46 \frac{J}{g*C} * m* (40 - 20) C

Solving:

300 J= 0.46 \frac{J}{g*C} * m* 20 C

m=\frac{300 J}{0.46 \frac{J}{g*C} * 20 C}

m= 32.61 g

4 0
3 years ago
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