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8_murik_8 [283]
3 years ago
12

All of the following are conditions in which the mantle will melt and liquefy EXCEPT ____________________________. a significant

amount of pressure is removed from the rock a significant amount of pressure is added to the mantle’s rock the temperature of the rock rises a significant amount of pressure is added to the mantle’s water
Chemistry
2 answers:
Tju [1.3M]3 years ago
5 0
The last thing doesn't make sense
Rama09 [41]3 years ago
3 0
The last one with water
it doesn't make sense.
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Gold is alloyed with other metals to increase its hardness in making jewelery.a) Consider a piece of gold jewelry the weighs 9.8
irinina [24]

Answer:

A). Percentage of gold by mass of the jewellery =

(6.059÷9.85) × 100 = 61.5%

B). Purity of the gold = 0.615 × 24 = 14.76 karat ~ 15 karat gold

Explanation:

Mass of jewellery = 9.85g Volume of jewellery = 0.675cm^3.

Density of gold = 19.3g/cm^3 and density of silver = 10.5g/cm^3

Let the volume of gold in the jewellery be X and the volume of silver in the jewellery be Y

Hence we have

Density = mass/volume or mass = volume × density = for gold = X × 19.3g/cm^3 and for the silver = Y × 10.5g/cm^3

19.3X +10.5Y = 9.85g

Also volume of jewellery is given by

Volume of silver in the jewellery + volume of gold in the jewellery = 0.675cm^3.

X + Y = 0.675cm^3.

Solving the above equations we have

Y = 0.675 - X

Which gives

19.3X + 10.5Y = 9.85g

19.3X + 10.5 × (0.675 - X) =9.85g

19.3X + 7.0875 - 10.5X = 9.85

8.8X + 7.0875 = 9.85

8.8X = 2.7625

or X = 0.3139 cm^3

But Y = 0.675 - X

Hence Y = 0.675 - 0.3139 = 0.3611 cm^3

Mass of gold in the jewellery = volume of gold × Density of gold = 0.3139 × 19.3 = 6.059 g

Also mass of silver = 10.5 × 0.3611 = 3.7913g

A). Percentage of gold by mass of the jewellery =

(6.059÷9.85) × 100 = 61.5%

B). Purity of the gold = 0.615 × 24 = 14.76 karat ~ 15 karat gold

7 0
3 years ago
An unknown acid solution has PH 3.4. 66% of the acid is ionized. Whats the pka?
juin [17]

Answer:

pKa=3.58

Explanation:

Hello,

In this case, since the pH defines the concentration of hydrogen:

pH=-log([H^+])

[H^+]=10^{-pH}=10^{-3.4}=3.98x10^{-4}

And the percent ionization is:

\% \ ionization=\frac{[H^+]}{[HA]}*100\%

We compute the concentration of the acid, HA:

[HA]=\frac{[H^+]}{\% \ ionization}*100\%=\frac{3.98x10^{-4}}{66\%}  *100\%\\\\

[HA]=6.03x10^{-4}

Thus, the Ka is:

Ka=\frac{[H^+][A^-]}{[HA]}=\frac{3.98x10^{-4}*3.98x10^{-4}}{6.03x10^{-4}}\\  \\Ka=2.63x10^{-4}

So the pKa is:

pKa=-log(Ka)=-log(2.63x10^{-4})\\\\pKa=3.58

Regards.

5 0
3 years ago
You run a “5k” for charity. How many feet do you run
Nikitich [7]

It would roughly be 16,404 feet. Or if you wanted to be more specific 16,404.2. I hope this helps.

5 0
3 years ago
Read 2 more answers
What is the pH of a solution that is 0.40 M NaBrO and 0.50 M HBrO (hypobromous acid) (Ka for HBrO = 2.3 x 10^-9)
Pie

Answer

pH=8.5414

Procedure

The Henderson–Hasselbalch equation relates the pH of a chemical solution of a weak acid to the numerical value of the acid dissociation constant, Kₐ. In this equation, [HA] and [A⁻] refer to the equilibrium concentrations of the conjugate acid-base pair used to create the buffer solution.

pH = pKa + log₁₀ ([A⁻] / [HA])

Where

pH = acidity of a buffer solution

pKa = negative logarithm of Ka

Ka =acid disassociation constant

[HA]= concentration of an acid

[A⁻]= concentration of conjugate base

First, calculate the pKa

pKa=-log₁₀(Ka)= 8.6383

Then use the equation to get the pH (in this case the acid is HBrO)

pH=8.6383+\log_{10}(\frac{0.40\text{ M}}{0.50\text{ M}})=8.5414

8 0
1 year ago
2.20 moles Sn to grams
mihalych1998 [28]
261.162 grams. Use the equation n=M÷Mr.
3 0
3 years ago
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