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3 years ago

6

What are two main groups of forces

1 answer:

Anuta_ua [19.1K]3 years ago

3 0

Answer:

There are 2 types of forces, contact forces and act at a distance force. Every day you are using forces. Force is basically push and pull. When you push and pull you are applying a force to an object.

Explanation:

You might be interested in

The elements in yellow are referred to as ___________.

Lemur [1.5K]

Answer:

Metalloids

Explanation:

hope this helps! :)

8 0

2 years ago

Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a

viva [34]

Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

$E^0_{[H^{+}/H_2]}=+0.00V$

$E^0_{[Ag^{+}/Ag]}=+0.799V$

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $H_2\rightarrow 2H^{+}+2e^-$

Reaction at cathode (reduction) : $Ag^{+}+1e^-\rightarrow Ag$

The balanced cell reaction will be,

$H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag$

Now we have to calculate the standard electrode potential of the cell.

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}$

$E^o_{cell}=(+0.799V)-(+0.00V)=+0.799V$

Therefore, the standard cell potential will be +0.799 V

4 0

3 years ago

The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K,

ludmilkaskok [199]

Answer: $4.3\times 10^{-13}s^{-1}$

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $600.0K$ = $6.1\times 10^{-8}s^{-1}$

$K_2$ = rate constant at $775.0$ = $?$

$Ea$ = activation energy for the reaction = 262 kJ/mol = 262000J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $600.0K$

$T_2$ = final temperature = $775.0K$

Now put all the given values in this formula, we get

$\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]$

$\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150$

$(\frac{6.1\times 10^{-8}}{K_2})=141253.8$

Therefore, the value of the rate constant at 775.0 K is $4.3\times 10^{-13}s^{-1}$

5 0

3 years ago

According to the following reaction, how many grams of hydrogen

Mila [183]

Answer:

7.03 g

Explanation:

Step 1: Write the balanced synthesis reaction

N₂(g) + 3 H₂(g) ⇒ 2 NH₃(g)

Step 2: Calculate the moles corresponding to 32.5 g of N₂

The molar mass of N₂ is 28.01 g/mol.

32.5 g × 1 mol/28.01 g = 1.16 mol

Step 3: Calculate the number of moles of H₂ needed to react with 1.16 moles of N₂

The molar ratio of N₂ to H₂ is 1:3. The moles of H₂ needed are 3/1 × 1.16 mol = 3.48 mol.

Step 4: Calculate the mass corresponding to 3.48 moles of H₂

The molar mass of H₂ is 2.02 g/mol.

3.48 mol × 2.02 g/mol = 7.03 g

8 0

3 years ago

How long did it take for a car to travel 20 meters with a speed of 10m/s

AfilCa [17]

Answer:

2 second

Explanation:

yes

6 0

2 years ago