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3 years ago

Foliated rocks are distinguished by ____.

Chemistry

2 answers:

Nataliya [291]3 years ago

4 0

Answer: Layers

Explanation:

kenny6666 [7]3 years ago

3 0

Foliated rocks are distinguished by layers, banding, or flakiness. Foliated rocks are given names depending their mineralogy and texture. These names can be slate, Phyllite and Schist. These rocks are formed within the Earth's interior under extremely high pressures that are unequal.

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An unknown liquid is composed of 34.31% c, 5.28% h, and 60.41% i. The molecular weight is 210.06 amu. What is the molecular form

ozzi

In an unknown liquid, the percentage composition with respect to carbon, hydrogen and iodine is 34.31%, 5.28% and 60.41% respectively.

Let the mass of liquid be 100 g thus, mass of carbon, hydrogen and oxygen will be 34.31 g, 5.28 g and 60.41 g respectively.

To calculate molecular formula of compound, convert mass into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of carbon, hydrogen and iodine is 12 g/mol, 1 g/mol and 126.90 g/mol.

Taking the ratio:

$C:H:I=n_{C}:n_{H}:n_{I}$

Putting the values,

$C:H:I=\frac{34.31 g}{12 g/mol}:\frac{5.28 g}{1 g/mol}:\frac{60.41 g}{126.90 g/mol}=6:11:1$

Thus, molecular formula of compound will be $C_{6}H_{11}I$ .

4 0

3 years ago

In the space provided, for each element, type the ionic charge of an atom with a full set of valence electrons. Then, type the n

kobusy [5.1K]

11. ionic charge +1, helium.

12. ionic charge 2-, neon.

13. ionic charge 3+, neon.

3 0

3 years ago

Consider the reaction below. Initially the concentration of SO2Cl2 is 0.1000 M. Solve for the equilibrium concentration of SO2Cl

tensa zangetsu [6.8K]

Answer:

$[SO_2Cl_2] = 0.09983 M$

Explanation:

Write the balance chemical equation ,

$SO_2Cl_2((g) = SO_2(g) + Cl_2(g)$

initial concenration of $SO_2Cl_2((g) =0.1M$

lets assume that degree of dissociation= $\alpha$

concenration of each component at equilibrium:

$[SO_2Cl_2] = 0.1-0.1\alpha$

$[SO_2] = 0.1\alpha$

$[Cl_2] = 0.1\alpha$

$Kc =\frac{0.1\alpha \times 0.1\alpha}{0.1-0.1\alpha}$

$Kc =\frac{0.1\alpha \times \alpha}{1-\alpha}$

as $\alpha$ is very small then we can neglect $1-\alpha$

therefore ,

$Kc ={0.1\alpha \times \alpha}$

$\alpha =\sqrt{\frac{Kc}{0.1}}$

$\alpha = 1.73 \times 10^{-3}$

Eqilibrium concenration of $[SO_2Cl_2] = 0.1-0.1\alpha = 0.1-0.1\times 0.00173$

$[SO_2Cl_2] = 0.09983 M$

4 0

3 years ago

The gas phase decomposition of sulfuryl chloride at 600 K SO2Cl2(g)SO2(g) + Cl2(g) is first order in SO2Cl2. During one experime

krok68 [10]

Answer: The rate constant for the reaction is $3.96\times 10^{-3}min^{-1}$

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample = 559 min

a = let initial amount of the reactant = $2.83\times 10^{-3}$

a - x = amount left after decay process = $3.06\times 10^{-4}$

$559min=\frac{2.303}{k}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}$

$k=\frac{2.303}{559}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}$

$k=3.96\times 10^{-3}min^{-1}$

The rate constant for the reaction is $3.96\times 10^{-3}min^{-1}$

6 0

4 years ago

If your front lawn is 18.0 feet wide and 20.0 feet long. And each square foot of lawn accumulates 1450 new snow flakes every min

creativ13 [48]

To solve this problem, we begin by first calculating the area of the front lawn. The length and width of the lawn was given and the area of a rectangle is given by the formula: Area = length x width. Thus, the area of the front lawn can be obtained by multiplying 18 ft by 20 ft, wherein we get 360 ft^2 as the area.

Second, the problem indicated that each square foot of lawn accumulates 1450 new snow flakes per minute. This can be translated into the expression 1450 snow flakes/ (minute·ft^2). In this way, we can convert it to units of mass (kg). Afterwards, we simply need to multiply it to the area of the lawn and convert minute to hour. The following expression is then used:

1450 snow flakes/ (minute·ft^2) x 1.90 mg/snow flake x 1 g/1000 mg x 1kg/1000 g x 360 ft^2 x 60 minutes/hour = 59.508 kg snow flake/hour

It is then calculated that 59.508 kg of snow flake accumulates in the lawn every hour.

6 0

3 years ago