All categories

3 years ago

Which would increase the reaction rate?

2 answers:

7 0

B) decreasing the reactant's particle size .

If a smaller reactant size was involved the reaction would occur faster as the quantity to partake in the reaction is smaller, and it would take lesser time compared to a bigger reactant.

Anna007 [38]3 years ago

6 0

B) decreasing the reactants particle size

You might be interested in

How to convert 0.000984 into scientific notation?

Olin [163]

9.84 * 10^-4
The easiest way I remember is making the decimal first and then seeing what way the 0's are. In this case it is going left so it will be a negative exponent and if you count from 9.84 and move the decimal left 4 that's your answer.

7 0

4 years ago

Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (

aleksley [76]

Answer: The amount of heat produced by the reaction is -21.36 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H_f_{(SO_3(g))}=-395.7kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2\times (-395.7))]-[(2\times (-296.8))+(1\times (0))]\\\\\Delta H_{rxn}=-197.8kJ/mol$

To calculate the number of moles, we use ideal gas equation, which is:

$PV=nRT$

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = $0.0831\text{ L. bar }mol^{-1}K^{-1}$

T = temperature of the mixture = $25^oC=[25+273]K=298K$

Putting values in above equation, we get:

$1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol$

To calculate the heat released of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

$-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ$

Hence, the amount of heat produced by the reaction is -21.36 kJ

3 0

3 years ago

Which of the graphs below might represent a mixture of pure water and ice exposed to a room temperature of 3°C?

Y_Kistochka [10]

Answer:

C. Graph C

Explanation:

We have a mixture of water and ice.

At 0 °C they are at equilibrium.

water-to-ice rate = ice-to-water rate

Next, we lower the temperature to -3 °C — just slightly below freezing.

The water will slowly turn to ice.

The water-to-ice rate will be slightly faster than the ice-to-water rate.

The purple bar will be slightly higher than the blue bar.

Graph C best represents the relative rates

A. is wrong. The ice-to-water rate is faster, so the water is melting. The temperature is slightly above freezing (say, 3 °C).

B. is wrong. The two rates are equal, so the temperature is 0 °C.

D. is wrong. The water-to-ice rate (freezing) is much greater than the ice-to-water rate, so the temperature is well below freezing( say, -10 °C).

6 0

4 years ago

Write the Hafferent modes by which diseases spread

Vinvika [58]

Answer:

Person to Person. When an infected person comes in contact with or exchanges body fluids with a non-infected person. ...
Droplet Transmission. ...
Spread by skin. ...
Spread through body fluids or blood. ...
Airborne Transmission. ...
Contaminated Objects. ...
Vector-Borne Diseases. ...
4.Food and Drinking Water.

Explanation:

6 0

2 years ago

Read 2 more answers

Which two electron configurations represent elements that would have similar chemical properties? (1) 1s22s22p4 (2) 1s22s22p5 (3

Anton [14]

Answer: Two electron configurations of elements that would have similar chemical properties are (2) and (4).

Explanation:

(1) $1s^22s^22p^4$

(2) $1s^22s^22p^5$

(3) $[Ar]4s^23d^5$

(4) $[Ar]4s^23d^{10}^4p^5$

Valence electrons : These are the electrons present in last principal quantum number of an atom of the element.

Two electron configurations represent elements that would have similar chemical properties are (2) and (4). This is because number of valence electrons present in both of them are same that is seven valence electrons . Also valence electrons of both the elements are present in p-orbital which means that they both belongs to same group in a periodic table. The members of same group in a periodic table have similar chemical properties.

3 0

3 years ago