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3 years ago

What are 4 examples of sediment that may become sedimentary rock

Chemistry

1 answer:

jeka943 years ago

7 0

Rockkkkkkkkkkkkkkkkskskskksks

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The value of the rate constant for a gas phase reaction can be changed by increasing the A. temperature of the reaction vessel.

prohojiy [21]

Answer:

temperature of the reaction vessel

Explanation:

temperature of the reaction vessel

3 0

2 years ago

What is the mass number of an ion with 106 electrons, 157 neutrons, and a +1 charge?

il63 [147K]

Answer:

264 g/mol

Explanation:

#electrons equal #protons = 106

Plus 1 charge => m protons = 106 + 1 = 107

Mass number: 107 + 157 = 264 g/mol

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2 years ago

PLEASE HELP ASAP!!!

solong [7]

The answer is, C: both Erin and Chris are correct.

8 0

3 years ago

Read 2 more answers

What is the type of bonding for ammonium lauryl sulfate?

DENIUS [597]

Answer:

Hydrogen bonding.

Explanation:

Ammonium lauryl sulfate is also known as ammonium dodecyl sulfate there are two parts in Ammonium lauryl sulfate one is nonpolar hydrocarbon and other part polar sulfate group.

Due to polarity of sulfate group its form hydrogen bond very easily.

It is mainly used as foaming agent the main reason of its use is very much soluble in water and making hydrogen bond with water.

7 0

3 years ago

For doping silicon with boron, silicon specimen was kept in gaseous atmosphere containing B2O3 that maintained the B concentrati

Kay [80]

Solution :

From Fick's law:

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=N_a$

Mass balance: Exits = Accumulation

$-N_A A = \frac{dm}{dt}$

$-N_A A = \frac{dVp}{dt}$

$-N_A A = \frac{dV}{dt}p$

$-N_A A = \frac{dhA}{dt}$

$-N_A A = \frac{dh}{dt} \times Ap$

From the last step, area cancels out and thus leaves :

$-N_A = \frac{dh}{dt} \times p$

So now we can substitute the $N_A$ by the Fick's law

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=\frac{dh}{dt} p$

Substituting the values we get

$=\frac{-4 \times 10^{-13}}{0.0001} \times (3 \times 10^{26} - C_{A2}) = \frac{dh}{dt} \times 2.46$

$=-4 \times 10^{-9} \times( 3 \times 10^{26} - C_{A2}) = 0.0001 \times 2.46$

$= -7800 \times 4 \times 10^{-9} \times (3 \times 10^{26}-C_{A2})=0.000246$ $=-(3 \times 10^{26}-C_{A2}) = 7.8846 \times 100 \times \frac{1}{69.62} \times 6.022 \times 10^{23}$

$C_{A2} = 6.85 \times 10^{28} \ \text{ boron atoms} /m^3$

3 0

2 years ago