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3 years ago

Electron pair geometry and molecular geometry may be the same or may be different.

Chemistry

1 answer:

soldier1979 [14.2K]3 years ago

8 0

True. For example, electron domain geometry and molecular geometry of water and ammonia are different.

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Advanced flow measuring devices ? <br> I need info about this topic please

Elenna [48]

Answer:

they are:-

venturi meter.

orfice plate.

Dall tube .

pitot tube.

Averaging pitot tube .

cone meter.

linear resis meter.

5 0

2 years ago

CHEMISTRY HELP<br> Once the following equation is balanced, what is the correct coefficient for Z₂?

yawa3891 [41]

Answer:

The coefficient of Z₂ is 1.

Explanation:

From the question given above:

X + ZY —> XY + Z₂

Next, we shall balance the equation to obtain the coefficient of Z₂. This can be obtained as follow:

X + ZY —> XY + Z₂

There is 1 atom of Z on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of ZY as shown below:

X + 2ZY —> XY + Z₂

There are 2 atoms of Y on the left side and 1 atom on the right side. It can be balance by putting 2 in front of XY as shown below:

X + 2ZY —> 2XY + Z₂

Now, we have 1 atom of X on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of X as shown below:

2X + 2ZY —> 2XY + Z₂

Now the equation is balanced.

Thus, the coefficient of Z₂ is 1.

8 0

3 years ago

The half-life of a radioactive isotope is the amount of time it takes for a quantity of that isotope to decay to one half of its

Sedaia [141]

Radio active decay reactions follow first order rate kinetics.

a) The half life and decay constant for radio active decay reactions are related by the equation:

$t_{\frac{1}{2}} =$ $\frac{ln 2}{k}$

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

Where k is the decay constant

b) Finding out the decay constant for the decay of C-14 isotope:

$Decay constant (k) = \frac{0.693}{t_{\frac{1}{2}}}$

$k = \frac{0.693}{5230 years}$

$k = 1.325 * 10^{-4} yr^{-1}$

c) Finding the age of the sample :

35 % of the radiocarbon is present currently.

The first order rate equation is,

$[A] = [A_{0}]e^{-kt}$

$\frac{[A]}{[A_{0}]} = e^{-kt}$

$\frac{35}{100} = e^{-(1.325 *10^{-4})t}$

$ln(0.35) = -(1.325 *10^{-4})(t)$

t = 7923 years

Therefore, age of the sample is 7923 years.

3 0

3 years ago

This is very urgent please hurry!!?!!??!

yawa3891 [41]

Answer:

The greatest acceleration when the unbalanced force is applied will be experienced in :

A) The box with a mass of 2 kg

Explanation:

According to second law of motion the external unbalanced force is directly proportional to rate of change of momentum.

F = (Final momentum - initial momentum)/time

Force is equal to the product of mass and acceleration

F = m x a

Here a= acceleration

m = mass of the object

If Force is constant then acceleration is inversely proportional to mass

$a =\frac{F}{m}$

A) The box with a mass of 2kg

F = 8 N

$a =\frac{8}{2}$

a = 4 m/s2

B) The box with the mass of 4kg

$a =\frac{8}{4}$

a = 2 m/s2

C) The box with a mass of 6kg

$a =\frac{8}{6}$

a = 1.33 m/s2

D) The box with a mass of 8kg

$a =\frac{8}{8}$

a = 1 m/s2

3 0

4 years ago

The 3 in 4p3 represents:

Marrrta [24]

B) the number of electrons

3 0

3 years ago

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