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3 years ago

9

What do we use electromagnet for

1 answer:

jekas [21]3 years ago

5 0

To make generators,doorlocks,etc

You might be interested in

The heisenberg uncertainty principle status that

Juli2301 [7.4K]

The Heisenberg uncertainty principle listed above states that it’s impossible to determine with high precision both the momentum and position of an electron simultaneously.

8 0

3 years ago

Calculate the ph of a buffer that is 0.225 m hc2h3o2 and 0.162 m kc2h3o2. the ka for hc2h3o2 is 1.8 Ã 10-5. 4.60 9.26 4.74 4.89

Stella [2.4K]

Answer:

The pH of the buffer solution is 4.60.

Explanation:

Concentration of acid = $[HC_2H_3O_2]=0.225 M$

Concentration of salt = $[KC_2H_3O_2]=0.162 M$

Dissociation constant = $K_a=1.8 \times 10^{-5}$

The pH of the buffer can be determined by Henderson-Hasselbalch equation:

$pH=pK_a+\log\frac{[salt]}{[acid]}$

$pH=-\log[1.8 \times 10^{-5}]+\log\frac{0.162 M}{0.225 M}$

pH = 4.60

The pH of the buffer solution is 4.60.

8 0

3 years ago

Read 2 more answers

Section I: Experimental Overview

Mariana [72]

Answer:

If you constructed graphs, what trends do they indicate in your data?

5 0

3 years ago

Be sure to answer all parts. Acetone is one of the most important solvents in organic chemistry. It is used to dissolve everythi

zimovet [89]

Answer:

a) 79.66 seconds is the half-life of the reaction.

b) It will take $4.776\times 10^1$ seconds for 34% of a sample of an acetone to decompose.

c) It will take $2.537\times 10^2$ seconds for 89% of a sample of an acetone to decompose.

Explanation:

The decomposition of acetone follows first order kinetics

The rate constant of the reaction = k = $8.7\times 10^{-3} s^{-1}$

a)

Half life of the reaction = $t_{1/2}$

For the first order kinetic half life is related to k by :

$t_{1/2}=\frac{0.693}{k}$

$t_{1/2}=\frac{0.693}{8.7\times 10^{-3} s^{-1}}=79.66 s=7.966\times 10^1 s$

79.66 seconds is the half-life of the reaction.

b)

Let the initial concentration of acetone be = $[A_o]$

Final concentration of acetone left after t time = [A]

$A=(100\%-34\%)[A_o]=66\%[A_o]=0.66[A_o]$

For the first order kinetic :

$[A]=[A_o]\times e^{-kt}$

$0.66[A_o]=[A_o]\times e^{-8.7\times 10^{-3} s^{-1}\times t}$

Solving for t;

$t=47.76 s$

It will take 47.76 seconds for 34% of a sample of an acetone to decompose.

c)

Let the initial concentration of acetone be = $[A_o]$

Final concentration of acetone left after t time = [A]

$A=(100\%-89\%)[A-o]=11\%[A_o]=0.11[A_o]$

For the first order kinetic :

$[A]=[A_o]\times e^{-kt}$

$0.11[A_o]=[A_o]\times e^{-8.7\times 10^{-3} s^{-1}\times t}$

Solving for t;

$t=2.537\times 10^2 s$

It will take $2.537\times 10^2$ seconds for 89% of a sample of an acetone to decompose.

3 0

3 years ago

What is the density of carbon dioxide gas at -25.2°C and 98.0 kPa? A. 0.232 g/L OB. 0.279 g/L OC. 0.994 g/L OD. 1.74 g/L O E. 2.

Black_prince [1.1K]

Answer:

E. $2.09\frac{g}{L}$

Explanation:

From the ideal gasses equation we have:

PV=nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

The number of moles is also expressed as: $n=\frac{mass}{Molar mass}$

If replacing this in the ideal gasses equation we have:

$PV=\frac{mass}{Molarmass}.RT$

If we pass V to divide, we have:

$P=\frac{mass}{V}.\frac{RT}{Molarmass}$

And the density d = $\frac{mass}{V}$ , so replacing, we have:

$P=\frac{dRT}{M}$

Solving for d, we have:

$d=\frac{P.M}{R.T}$

Now we have to be sure that we have the correct units, so we need to convert the units for pressure and temperature:

-Convert P=98kPa to atm

$98.0kPa*\frac{0.00986923atm}{1kPa}=0.97atm$

-Convert T=-25.2°C to K

$-25.2^{o}C+273.15=247.95K$

Finally we can replace the values in the equation:

$d=\frac{(0.97atm)*(44.01\frac{g}{mol})}{(0.082\frac{atm.L}{mol.K})*(247.15K)}$

$d=2.09\frac{g}{L}$

8 0

3 years ago