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3 years ago

Need help please ASAP !!!!

Chemistry

1 answer:

Dafna11 [192]3 years ago

4 0

Answer:

ur down bad, I forgot how to do

Explanation:

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What is the pOH of a 2.6 x 10-6 M H+ solution?

melomori [17]

Answer:

Approximately $8.41$ (assuming that the solution is at $\rm 25^\circ C$ , under which $K_{\rm w} = 10^{-14}$ .)

Explanation:

Let ${\rm [H^{+}]}$ and ${\rm [OH^{-}]}$ denote the concentration of $\rm H^{+}$ and $\rm OH^{-}$ respectively.

Let $K_{\rm w}$ denote the self-ionization constant of water. The exact value of $K_{\rm w}\!$ depends on the temperature of the solution. $K_{\rm w} =10^{-14}$ at $\rm 25^\circ C$ .

The product of ${\rm [H^{+}]}$ and ${\rm [OH^{-}]}$ in a solution (with $\rm M$ , or moles per liter, as the unit) is supposed to be equal to the $K_{\rm w}$ value of that solution at the corresponding temperature. In other words:

${\rm [H^{+}]} \cdot {\rm [OH^{-}]} = K_{\rm w}$ .

Rearrange to obtain an expression for ${[\rm OH^{-}]}$ :

$\begin{aligned}{\rm [OH^{-}]} &= \frac{K_{\rm w}}{[\rm H^{+}]}\end{aligned}$ .

Assume that the solution in this question is at $\rm 25^\circ C$ (for which $K_{\rm w} =10^{-14}$ .) For this solution:

$\begin{aligned}{\rm [OH^{-}]} &= \frac{K_{\rm w}}{[\rm H^{+}]} \\ &= \frac{10^{-14}}{2.6 \times 10^{-6}}\approx 3.85\times 10^{-9}\; \rm M\end{aligned}$ .

Hence, the $\rm pOH$ of this solution would be:

$\begin{aligned}\rm pOH &= -\log_{10}{\rm [OH^{-}]} \\&\approx -\log_{10} (3.85 \times 10^{-9}) \approx 8.41 \end{aligned}$ .

3 0

3 years ago

2 C6H14 + 19 O2 --> 12 CO2 + 14 H2O

vivado [14]

Answer:

1.27 moles (3 s.f.)

Explanation:

Mole ratio of water: C6H14

= 14:2

= 7:1

This means that to produce 7 moles of water, 1 mole of C6H14 is needed. Or 1/7 mole of C6H14 is needed to produce 1 mole of water. So if you need 8.86 moles of water, 8.86(1/7) moles of C6H14 is needed.

For 8.86mol of water, moles of C6H14 needed

= 8.86/7

= 1.27 moles (3 s.f.)

4 0

4 years ago

By referring to particles, explain why the white ring formed in the glass tube

Leya [2.2K]

Answer:

Ammonia and Hydrogen Chloride.

Explanation:

Assuming this is what you're referring to, Ammonia (NH3) and Hydrogen Chloride(NCI3). The concentrated ammonia (NH3) is placed on a pad in one end of a tube and hydrochloric acid (NCI3) on a pad at the other. Shortly after the gases will begin diffusing far enough to meet, a ring of solid ammonium chloride (NH4Cl) will be formed.

7 0

3 years ago

Rick uses an electronic air sampler to determine the amount of dangerous chemicals in the air.

lesya692 [45]

Answer:

The first three options given out of the five options are correct.

It collects chemical data. 
It records real-time sampling data. 
It processes and stores chemical data.

Explanation:

An electronic air sampler is a device that is used by people to check apprehension of contamination having individuals with the help of certain volume of air sampling.

The Device is used for recording the chemical data of contamination and the real-time data is sampled and recorded and the data is processed and stored. The air sampling is done in two ways.

3 0

3 years ago

(science)

ra1l [238]

Answer:

b. . the element symbol

6 0

3 years ago

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