What part of the autonomic

what information can a foliated metamorphic rock provide you about the conditions under which it formed

uysha [10]

The information that a foliated metamorphic rock provide about the conditions it was formed under are:(for many diffrent rocks like Phyllite and Amphibolite)
you can tell by its features like if its smooth and sleek or if its rough and bumpy those physical features.

3 0

3 years ago

In The Lord of the Rings, Frodo and Sam use a rope to rappel down a cliff. The rope was 30 ells long. How high was the cliff in

Tcecarenko [31]

Answer:

324 meters

Explanation:

I don't really have an explanation

7 0

3 years ago

Acenapthalene has the empirical formula C6H5. A solution of 0.515 g of acenapthalene in 15.0 g CHCl3 boils at 62.5oC. The normal

german

Answer:

The molecular formula of an ascenapthalene is $C_{12}H_{10}$

Explanation:

$\Delta T_b=K_b\times m$

$\Delta T_b=K_b\times \frac{\text{Mass of acenapthalene}}{\text{Molar mass of acenapthalene}\times \text{Mass of chloroform in Kg}}$

where,

$\Delta T_f$ =Elevation in boiling point = $(62.5-61.7)^oC=0.8^oC$

Mass of acenapthalene = 0.515 g

Mass of $CHCl_3$ = 15.0 g = 0.015 kg (1 kg = 1000 g)

$K_b$ = boiling point constant = 3.63 °C/m

m = molality

Now put all the given values in this formula, we get

$0.8^0C=3.67 ^oC/m\times \frac{0.515}{\text{Molar mass of acenapthalene}\times 0.015kg}$

$\text{Molar mass of acenapthalene}=155.7875 g/mol$

Let the molecule formula of the Acenapthalene be $C_{6n]H_{5n}$

$6n\times 12 g/mol+5n\times 1 g/mol=155.7875 g/mol$

n = 2.0

The molecular formula of an ascenapthalene is $C_{12}H_{10}$

4 0

3 years ago

Give 10 chemical properties of common polymers

algol [13]

Answer:

Some of the useful properties of various engineering polymers are high strength or modulus to weight ratios (light weight but comparatively stiff and strong), toughness, resilience, resistance to corrosion, lack of conductivity (heat and electrical), color, transparency, processing, and low cost

Explanation:

3 0

2 years ago

Read 2 more answers

What is the molariity of a 50.0 mL aqueous solution containing 10.0 grams of table salt, NaCl?

tia_tia [17]

Answer:

3 mol/L

Explanation:

You should know or have the equation to solve for Molarity which is;

M = n/v (M: Molarity) (n: moles of solute) (v: Liters of solute)

You can start off differently but I would start by converting the mL to L. This is your "v" value.

50.0 mL/ 1000 mL = 0.05 L

Now, you have to convert grams to moles in order to solve for molarity (M).

1.) On the periodic table find the molecular weights of Na and Cl.

Na= 22.99 g/mol Cl= 35.45g/mol

2.) Add them together to have their combined molecular weights.

22.99 + 35.45= 58.44 g/mol

3.) Now, you're going to use the "picket fence method" or whichever your teacher taught you to convert from grams to moles. This will be your "n" value. (I cannot show it on here without it looking weird, so my sincere apologies.)

10.0 g/ 58.44 g = 0.17111 mol

4.)You are now going to plug in your answers into the equation for Molarity.

M= 0.17111 mol / 0.05 L = 3.4222

5.) I am sure your professor might be a stickler so for sig figs sake when you multiply or divide use the smallest amount of sig figs you see which is 1. Round 3.4222 to 3 mol/L

Sorry this explanation is long let me know if you need a better more written out sample.

3 0

3 years ago