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2 years ago

12

What type of bond would form between two atoms of selenium?

1 answer:

siniylev [52]2 years ago

8 0

Answer:

the correct answer is B

hope it helps !!!

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A thermometer is placed in a beaker full of water that has been sitting on the lab table for a few hours. The beaker is placed o

krek1111 [17]

Based on the information I would assume B, 73 degrees...

It shouldn't be A, 4 minutes on the burner should increase the temperature.

If it were D, it would be beyond boiling, and water takes a decent amount of energy to heat, D should be all vapor.

Same logic for C, it's basically almost boiling.

I would say 73 degrees seems most reasonable for 4 minutes.

7 0

3 years ago

Read 2 more answers

The rate constant for a certain reaction is k = 5.40×10−3 s−1 . If the initial reactant concentration was 0.100 M, what will the

IceJOKER [234]

Answer : The concentration after 17.0 minutes will be, $4.05\times 10^{-4}M$

Explanation :

The expression for first order reaction is:

$[C_t]=[C_o]e^{-kt}$

where,

$[C_t]$ = concentration at time 't' (final) = ?

$[C_o]$ = concentration at time '0' (initial) = 0.100 M

k = rate constant = $5.40\times 10^{-3}s^{-1}$

t = time = 17.0 min = 1020 s (1 min = 60 s)

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

$[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}$

$[C_t]=4.05\times 10^{-4}M$

Thus, the concentration after 17.0 minutes will be, $4.05\times 10^{-4}M$

4 0

2 years ago

Plz Help I have One more time To retake and I can't fail it..

navik [9.2K]

Answer:

A

Explanation:

I think the answer would be A

4 0

2 years ago

What is the Ka of a weak acid (HA) if the initial concentration of weak acid is 4.5 x 10-4 M and the pH is 6.87? (pick one)

rewona [7]

Answer:

Ka = $4.04 \times 10^{-11}$

Explanation:

Initial concentration of weak acid = $4.5 \times 10^{-4}\ M$

pH = 6.87

$pH = -log[H^+]$

$[H^+]=10^{-pH}$

$[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M$

HA dissociated as:

$HA \leftrightharpoons H^+ + A^{-}$

(0.00045 - x) x x

[HA] at equilibrium = (0.00045 - x) M

x = $1.35 \times 10^{-7}\ M$

$Ka = \frac{[H^+][A^{-}]}{[HA]}$

$Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}$

0.000000135 <<< 0.00045

$Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}$

5 0

2 years ago

Bond [772]

The answer is A. The pH decreases

7 0

2 years ago