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3 years ago

How do river systems, watersheds, and divides interact?

Chemistry

2 answers:

love history [14]3 years ago

5 0

Answer:

They all end up in the oceon

Explanation:

Mnenie [13.5K]3 years ago

5 0

Answer:

Explanation:

i think they will all end up in ocean

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How many moles of chlorine are in 6.67x10^40 chlorine molecules?

sweet-ann [11.9K]

To solve for the number of moles, we simply have to use the Avogadros number which states that there are 6.022 x 10^23 molecules per mole. Therefore:

number of moles = 6.67 X 10^40 chlorine molecules / (6.022 x 10^23 molecules / mole)
number of moles = 1.108 x 10^17 moles

4 0

3 years ago

Read 2 more answers

Calculate the reaction quotient Qp for the following redox reaction: 14H+ + Cr2O72- + 6Cl- ----> 2Cr3+ + 3Cl2 + 7H2O The reac

stich3 [128]

Answer:

Value of $Q_{p}$ for the given redox reaction is $1.0\times 10^{-8}$

Explanation:

Redox reaction with states of species:

$14H^{+}(aq.)+Cr_{2}O_{7}^{2-}(aq.)+6Cl^{-}(aq.)\rightarrow 2Cr^{3+}(aq.)+3Cl_{2}(g)+7H_{2}O(l)$

Reaction quotient for this redox reaction:

$Q_{p}=\frac{[Cr^{3+}]^{2}.P_{Cl_{2}}^{3}}{[H^{+}]^{14}.[Cr_{2}O_{7}^{2-}].[Cl^{-}]^{6}}$

Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of $H_{2}O$ is taken as 1 due to the fact that $H_{2}O$ is a pure liquid.

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

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

Plug in all the given values in the equation of $Q_{p}$ :

$Q_{p}=\frac{(0.10)^{2}\times (0.010)^{3}}{(10^{-0.0})^{14}\times (1.0)\times (1.0)^{6}}=1.0\times 10^{-8}$

7 0

3 years ago

All of the following methods would increase the solubility of a solid solute, except for

cricket20 [7]

May i please have a(n) answer choices please because it would be a lot better if it was like that and then ill answer it

8 0

3 years ago

How did he know that the nucleus was positively charged?

Harman [31]

'cause alphe-particle which was +ve charge, get repulsion from the atom, so he deducted that.......

8 0

3 years ago

An alkyne with the molecular formula C5H8 was reduced with H2 and Lindlar's catalyst. Hydroboration-oxidation of the resulting a

irinina [24]

Since the addition of the H2O in the last step of hydroboration is anti-Markovnikov, the starting material is 1-pentyne.

The addition of H2 to C5H8 yields an alkene when a Lindlar catalyst is used. Recall that the Lindlar catalysts poisons the process so that the addition do not go on to produce an alkane.

When hydroboration is carried out on the alkene, we are told that a primary alcohol was obtained. We must note that in the last step of hydroboration, water is added in an anti- Markovnikov manner to yield the primary alcohol. Hence, the starting material must be 1-pentyne as shown in the image attached.

Learn more: brainly.com/question/2510654

6 0

3 years ago