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tekilochka [14]
3 years ago
5

Which of the following conditions increases the frequency of collisions without changing the energy or concentration of reactant

s?
A: increasing the surface area of the substance
B: adding more reactant molecules
C: increasing the temperature
D: increasing the pressure
Chemistry
2 answers:
Brums [2.3K]3 years ago
5 0
You can promote collisions between molecules when you increase their kinetic energy by either increasing the temperature or pressure conditions of the systems. However, it is indicated that you must not change the energy. So, we eliminate choices C and D. Choice B is contradictory to the limitation that you do not increase the concentration of the reactants. Technically, all of the choices promote frequent collisions. But due to the limitations set, the accepted answer is letter A. When you increase the surface area by using powdered reactants instead of cubes, you increase the areas for susceptibility of collisions.
ladessa [460]3 years ago
3 0

Answer:

A: increasing the surface area of the substance

Explanation:

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Shtirlitz [24]

Answer:

in 4 liters of solution

MgSO4) are dissolved in 0.965 L of a 0.0575 M solution? Molar Mass Mg: 24.30 g/mol Molar Mass S: 32.06 g/mol Molar Mass O: 16.00 g/

8 0
2 years ago
Hi, someone knows how to balance the following equation, and if you can thank you if you can explain how that is done:
kari74 [83]
To begin with, the equation given is not correct.
Correct equation is : CaCO3 + HCl ---> CaCl2 + H2O + CO2
It's CaCl2 not CaCl because Ca has a valency of 2 

    LHS                          RHS
CaCO3 + HCl ---> CaCl2 + H2O + CO2
First of all, to balance the equation you must look at the number of atoms on each side of the equation. 
we have 2 H on the RHS and 1 H on the LHS. So, we put a 2 on the LHS

CaCO3 + 2HCl ---> CaCl2 + H2O + CO2
Check for the LHS: 1 Ca, 1 C, 3 O, 2 H & 2 Cl on the LHS
Now check for the RHS: 1 Ca, 2 Cl, 2 H, 1 C & 3 O

Hope it helped!
7 0
3 years ago
How much energy is released by the decay of 3 grams of 230Th in the following reaction 230 Th - 226Ra + "He (230 Th = 229.9837 g
Serhud [2]

<u>Answer:</u> The energy released for the decay of 3 grams of 230-Thorium is 2.728\times 10^{-15}J

<u>Explanation:</u>

First we have to calculate the mass defect (\Delta m).

The equation for the alpha decay of thorium nucleus follows:

_{90}^{230}\textrm{Th}\rightarrow _{88}^{226}\textrm{Ra}+_2^{4}\textrm{He}

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

\Delta m=(m_{Ra}+m_{He})-(m_{Th})

\Delta m=(225.9771+4.008)-(229.9837)=1.4\times 10^{-3}amu=2.324\times 10^{-30}kg

(Conversion factor: 1amu=1.66\times 10^{-27}kg )

To calculate the energy released, we use Einstein equation, which is:

E=\Delta mc^2

E=(2.324\times 10^{-30}kg)\times (3\times 10^8m/s)^2

E=2.0916\times 10^{-13}J

The energy released for 230 grams of decay of thorium is 2.0916\times 10^{-13}J

We need to calculate the energy released for the decay of 3 grams of thorium. By applying unitary method, we get:

As, 230 grams of Th release energy of = 2.0916\times 10^{-13}J

Then, 3 grams of Th will release energy of = \frac{2.0916\times 10^{-13}}{230}\times 3=2.728\times 10^{-15}J

Hence, the energy released for the decay of 3 grams of 230-Thorium is 2.728\times 10^{-15}J

5 0
3 years ago
Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules. Several other gases make up the remaining 1% of air molecules.
kherson [118]

Answer:

The partial pressure of the other gases is 0.009 atm

Explanation:

Step 1: Data given

Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules and 1% of other gases.

The atmospheric pressure = 0.90 atm

Step 2: Calculate mol fraction

If wehave  100  moles of air, 78 moles will be nitrogen,

21  moles will be  oxygen, and  1  mol will be other gases.

Mol fraction = 1/100 = 0.01

Step 3: Calculate the partial pressure of the other gases

Pgas = Xgas * Ptotal

⇒ Pgas = the partial pressure = ?

⇒ Xgas = the mol fraction of the gas = 0.01

⇒Ptotal = the total pressure of the pressure = 0.90 atm

Pgas = 0.01 * 0.90 atm

Pgas = 0.009 atm

The partial pressure of the other gases is 0.009 atm

4 0
3 years ago
The gas in the piston is at constant temperature. A student increases the pressure on the piston from 2 atm to 3 atm. The observ
Vesna [10]

Answer: row 1, the volume decreases when the pressure increased

Explanation:

7 0
3 years ago
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