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3 years ago

PLZZ HELP ME ASAP

Chemistry

2 answers:

Amiraneli [1.4K]3 years ago

5 0

The answer is A according to my teacher

galben [10]3 years ago

3 0

I think it is yes, the reaction may require an activation energy

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The _____________ is the temperature at which a substance changes from solid to liquid; ________________ is the temperature at w

Aleksandr-060686 [28]

The _____melting point________ is the temperature at which a substance changes from solid to liquid; _______boiling point_________ is the temperature at which a substance changes from a liquid to as gas; _______vapourisation_________ is the process by which atoms of molecules leave a liquid and become a gas.

5 0

4 years ago

Select all possible answers:

Dafna11 [192]

Answer:

c and d are correct

Explanation:

In A, false because in Valence Electrons, the more the valences, the more stable an atom is.

In B, false because atoms cannot readily gain or lose valence electrons as the number of valence electrons is determined by the column they are in.

In C, true because the more the valence electrons, the more the stability of an atom.

In D, true as electron placing is important and the reactivity of an atom is important.

So C and D are true!

7 0

3 years ago

Read 2 more answers

Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given

tekilochka [14]

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

$K_a$ = $1.34*10^{-5$

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝) $\alpha = \sqrt{\frac{K_a}{C} }$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }$

$\alpha = \sqrt{4.9629*10^{-5}}$

$\alpha = 7.044*10^{-3$

percentage% (∝) = $7.044*10^{-3}*100$

= 0.704%

For Part B; where Concentration of B = $7.84*10^{-2$ M

$\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }$

$\alpha = \sqrt{1.709*10^{-4} }$

$\alpha = 0.0130\\$

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= $1.92*10^{-2} M$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }$

$\alpha = \sqrt{6.802*10^{-4}}$

$\alpha =0.02608$

percentage% (∝) = 0.02608 × 100%

= 2.60%

7 0

3 years ago

Please help a s a p!

ValentinkaMS [17]

Predator 1: wolves
Predator 2: foxes
Prey 1: bison
Prey 2: prairie dogs

3 0

3 years ago

A rigid tank contains 1.40 moles of an ideal gas. Determine the number of moles of gas that must be withdrawn from the tank to l

sergey [27]

Answer : The final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.

Explanation :

As we know that:

$PV=nRT$

At constant volume and temperature of gas, the pressure will be directly proportional to the number of moles of gas.

The relation between pressure and number of moles of gas will be:

$\frac{P_1}{P_2}=\frac{n_1}{n_2}$

where,

$P_1$ = initial pressure of gas = 24.5 atm

$P_2$ = final pressure of gas = 5.30 atm

$n_1$ = initial number of moles of gas = 1.40 moles

$n_2$ = final number of moles of gas = ?

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

$\frac{24.5atm}{5.30atm}=\frac{1.40mol}{n_2}$

$n_2=0.301mol$

Therefore, the final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.

8 0

3 years ago