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3 years ago

The Earth's asthenosphere is a structural layer made up on what?

Chemistry

2 answers:

Vera_Pavlovna [14]3 years ago

8 0

It is made up of gas

ehidna [41]3 years ago

5 0

Gas is the answer ok

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According to the usda food standards and labeling, vinegar must contain at least 4% (m/m) acetic acid. does your vinegar sample

Anastasy [175]

Answer:

A for me

Explanation:

I dont know what you mean by this as this is an opinon question

7 0

2 years ago

When I cut it a gas will irritate my eye. Sometimes when I'm cooking you might see me cry.

Elena-2011 [213]

Lol, onions. Onions produce a chemical irritant called synpropanerhial-S-oxide.

3 0

3 years ago

What is the pH of this solution?

Vesnalui [34]

Answer:

pH = 11.216.

Explanation:

Hello there!

In this case, according to the ionization of ammonia in aqueous solution:

$NH_3+H_2O\rightleftharpoons NH_4^++OH^-$

We can set up its equilibrium expression in terms of x as the reaction extent equal to the concentration of each product at equilibrium:

$Kb=\frac{[NH_4^+][OH^-]}{[NH_3]} \\\\1.80x10^{-5}=\frac{x*x}{0.150-x}$

However, since Kb<<<1 we can neglect the x on bottom and easily compute it via:

$1.80x10^{-5}=\frac{x*x}{0.150}\\\\x=\sqrt{1.80x10^{-5}*0.150}=1.643x10^{-3}M$

Which is also:

$[OH^-]=1.643x10^{-3}M$

Thereafter we can compute the pOH first:

$pOH=-log(1.643x10^{-3}M)\\\\pOH=2.784$

Finally, the pH turns out:

$pH=14-2.784\\\\pH=11.216$

Regards!

5 0

3 years ago

A solution was prepared by dissolving 2.2 g of an unknown solute in 16.7 g of CCl4. A thermal analysis was performed for this so

laila [671]

Answer:

Molar mass of unknown solute is 679 g/mol

Explanation:

Let us assume that the solute is a non-electrolyte.

For a solution with non-electrolyte solute remains dissolved in it -

Depression in freezing point of solution, $\Delta T_{f}=K_{f}.m$

where, m is molality of solute in solution and $K_{f}$ is cryogenoscopic constant of solvent.

Here $\Delta T_{f}=(-22.9^{0}\textrm{C})-(-28.7^{0}\textrm{C})=5.8^{0}\textrm{C}$

If molar mass of unknown solute is M g/mol then-

$m=\frac{\frac{2.2}{M}}{0.0167}mol/kg$

So, $5.8^{0}\textrm{C}=29.9^{0}\textrm{C}/(mol/kg)\times \frac{\frac{2.2g}{M}}{0.0167}mol/kg$

so, M = 679 g/mol

4 0

3 years ago

Naming Organic compounds

Oduvanchick [21]

i don't know but try putting this diagram into a question on google. you should be able to get some type of answer if not the right answer. good luck and * hint* you can make a really good question out of the sentence on top of the DIAGRAM. I hope this was helpful. please let me know in the comments: )

7 0

4 years ago