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Marta_Voda [28]
3 years ago
6

What most likely to cuase the rate of a chemical reaction to increase

Chemistry
2 answers:
-BARSIC- [3]3 years ago
7 0

Answer: adding a catalyst or increasing temperature

Explanation:

tensa zangetsu [6.8K]3 years ago
3 0

Answer:

An increase on temperature.

Explanation:

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skad [1K]
Question 5)

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Every period of time, the radioactive isotopes halve in numbers as they emit radiation.

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6 0
3 years ago
Consider the above reaction. If some CO2 were removed from the container, what would the reaction do in an attempt to replace th
faltersainse [42]

Answer:

See explanation

Explanation:

The reaction to be considered is shown below;

H2CO3<------->CO2 + H2O

We know  that when a constraint such as a sudden change in concentration, pressure or temperature is imposed on  a reaction system in equilibrium, the system has to adjust itself by shifting in a particular direction in order to cancel the constraint.

Now, if we remove CO2, the equilibrium position must shift to the right by the decomposition of more H2CO3 to establish equilibrium again.

4 0
3 years ago
50cm3 of 1 mol/dm3 HCl at 30°C was mixed with 50cm3 of 1mol/dm3 NaOH at 30°C in a styrofoam calorimeter. The temperature of the
trapecia [35]

Answer:

-21 kJ·mol⁻¹  

Explanation:

Data:

                    H₃O⁺ +  OH⁻ ⟶ 2H₂O

       V/mL:    50         50  

c/mol·dm⁻³:   1.0         1.0

     

ΔT = 4.5 °C  

       C = 4.184 J·°C⁻¹g⁻¹

C_cal = 50 J·°C⁻¹

Calculations:

(a) Moles of acid

\text{Moles of acid} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}\\\\\text{Moles of base} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}

So, we have 0.050 mol of reaction

(b) Volume of solution

V = 50 dm³ + 50 dm³ = 100 dm³

(c) Mass of solution

\text{Mass of solution} = \text{100 dm}^{3} \times \dfrac{\text{1.00 g}}{\text{1 dm}^{3}} = \text{100 g}

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

q₁ + q₂ + q₃ = 0

     nΔH   +         mCΔT       + C_calΔT = 0

0.050ΔH + 100×4.184×4.5 +   50×4.5  = 0

0.050ΔH +          1883        +      225    = 0

                                  0.050ΔH + 2108 = 0

                                              0.050ΔH = -2108

                                                        ΔH = -2108/0.0500

                                                              = -42 000 J/mol

                                                              = -42 kJ/mol

This is the heat of reaction for the formation of 2 mol of water

The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.

5 0
3 years ago
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