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r-ruslan [8.4K]
3 years ago
9

What is a common land form that is formed when chemical weathering, specifically carbonation, is taking place?

Chemistry
1 answer:
skad [1K]3 years ago
6 0

Answer:

plateaus

Explanation:

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How many moles are in 3.4 x 10-7 grams of Silicon dioxide, SiO2?
Varvara68 [4.7K]

Answer:

Number of moles = 0.057 × 10⁻⁷  mol

Explanation:

Given data:

Mass of SiO₂ = 3.4 × 10⁻⁷ g

Number of moles = ?

Solution:

Number of moles = mass/molar mass

Molar mass of SiO₂ = 60 g/mol

by putting values,

Number of moles =  3.4 × 10⁻⁷ g / 60 g/mol

Number of moles = 0.057 × 10⁻⁷  mol

8 0
3 years ago
Which description best characterization the motion of particles in a solid
marta [7]

Are there any choices? Because from what the question is it seems like we need choices to help

4 0
3 years ago
Read 2 more answers
Calculate the pH of a solution prepared by mixing: (Show your work for these calculations) pk of acetic acid is 4.75 a. two mole
kupik [55]

Answer:

Explanation:

To calculate pH you need to use Henderson-Hasselbalch formula:

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

Where HA is the acid concentration and A⁻ is the conjugate base concentration.

The equilibrium of acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75

Where <em>CH₃COOH </em>is the acid and <em>CH₃COO⁻ </em>is the conjugate base.

Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:

pH = 4,75 + log₁₀ \frac{[CH_{3}COO^-]}{[CH_{3}COOH]}

a) The pH is:

pH = 4,75 + log₁₀ \frac{[2 mol]}{[2 mol]}

<em>pH = 4,75</em>

<em></em>

b) The pH is:

pH = 4,75 + log₁₀ \frac{[2 mol]}{[1mol]}

<em>pH = 5,05</em>

<em></em>

I hope it helps!

7 0
3 years ago
What is the molarity of Li in a solution that contains of 46.552ppm lithium ferrocyanide Li3fe(cn)6
malfutka [58]

Answer:

hgfdjsuejssj_uyghgjiyr656⁸8⁴83jbv

3 0
2 years ago
Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s
NeX [460]

Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

¡Saludos!

3 0
3 years ago
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