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wlad13 [49]
3 years ago
9

4HCI + O2 =2H20 + Cl2

Chemistry
1 answer:
Sergio039 [100]3 years ago
6 0

The rate of the backward reaction increases

Explanation:

It is evident that if the reaction is left to proceed spontaneously, the forward reaction is favored because it results in a decrease in pressure in the system (The total reactants have 5 moles and the products have 3 in total).

Increasing H₂O concentration is then reaction, therefore, stymies the forward reaction and favors the reserves reaction. This is because the reverse reaction will lead to reduced pressure.

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How do you balance this Chemical Equation? <br>C5H12O + O2 = CO2 + H2O​
ICE Princess25 [194]

Answer:

The chemical equation by putting, a 2 on C₅H₁₂O, 15 on O₂, 10 on CO₂ , and 12 on H₂O in the equation;

2C₅H₁₂O + 15O₂ → 10CO₂ + 12H₂O​

Explanation:

  • Chemical equations are balanced by putting coefficients on the reactants and products to ensure the total number of atoms on the left side equal to those on the right side.
  • Balancing chemical equations is done to make chemical equations obey the law of conservation of mass.
  • According to the law of conservation of mass, the mass of the reactants should always be equal to the mass of products.
  • This is done by balancing chemical equations to ensure the total number of atoms on the left side is equal to that on the right side.
  • Therefore, the balanced equation is;

          2C₅H₁₂O + 15O₂ → 10CO₂ + 12H₂O​

3 0
3 years ago
How many grams in a kilogram,how many quarts are in a pint and if how many ml of h20 = 1g
vodomira [7]

Answer:

1 g  +  10 g

Explanation:

5 0
3 years ago
Read 2 more answers
in 1866, Gregor Mendel published his paper that described the research that led to his discovery of the laws of inheritance. Mor
julia-pushkina [17]
The answers will help you!!!!!
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The statement describes from the relationship between multiple approaches and familiarity with similar work. The answers is would be last sentences. D. Multiple approaches may occur because scientists develop similar interests independently.
4 0
3 years ago
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HEEEEELP ASAPP IM AWARDING 30 points!!!!
alexira [117]

Answer:

\rm S^{2-}.

Explanation:

Based on the electron configuration of this ion, count the number of electrons in this ion in total:

2 + (2 + 6) + (2 + 6) = 18.

Each electron has a charge of (-1).

Atoms are neutral and have 0 charge. However, when an atom gains one extra electron, it becomes an ion with a charge of (-1). Likewise, when that ion gains another electron, the charge on this ion would become (-2).

The ion in this question has a charge of (-2). In other words, this ion is formed after its corresponding atom gains two extra electrons. This ion has 18 electrons in total. Therefore, the atom would have initially contained 18 - 2 = 16 electrons. The atomic number of this atom would be 16.

Refer to a modern copy of the periodic table. The element with an atomic number of 16 is sulphur with atomic symbol \rm S. To denote the ion, place the charge written backwards ("2-" for a charge of (-2)) as the superscript of the atomic symbol:

\rm S^{2-}.

5 0
3 years ago
Read 2 more answers
The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C
avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

q=c\times (T_{final}-T_{initial})

where,

q = heat gained = ?

c = specific heat = 5.20kJ/^oC

T_{final} = final temperature = 27.43^oC

T_{initial} = initial temperature = 22.93^oC

Now put all the given values in the above formula, we get:

q=5.20kJ/^oC\times (27.43-22.93)^oC

q=23.4kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = \frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole

\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

\Delta H=\Delta U+\Delta n_gRT

or,

\Delta U=\Delta H-\Delta n_gRT

where,

\Delta H = change in enthalpy = -2805.8kJ/mol

\Delta U = change in internal energy = ?

\Delta n_g = change in moles = 0   (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = 27.43^oC=273+27.43=300.43K

Now put all the given values in the above formula, we get:

\Delta U=\Delta H-\Delta n_gRT

\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K

\Delta U=-2805.8kJ/mol-0

\Delta U=-2805.8kJ/mol

Therefore, the internal energy change is -2805.8 kJ/mol

5 0
3 years ago
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