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3 years ago

The layering of eroded sediments is called:

Chemistry

1 answer:

Aleks [24]3 years ago

6 0

4. the rock cycle is the layering of eroded sediments

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Chlorine (Cl) is 76% chlorine-35 and 24% chlorine-37. Determine the average atomic mass of chlorine. Show/Explain work.

Flura [38]

Answer:

Average atomic mass of chlorine is 35.48 amu.

Explanation:

Given data:

Percent abundance of Cl-35 = 76%

Percent abundance of Cl-37 = 24%

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100

Average atomic mass = (76×35)+(24×37) /100

Average atomic mass = 2660 + 888 / 100

Average atomic mass = 3548/ 100

Average atomic mass = 35.48 amu

Average atomic mass of chlorine is 35.48 amu.

4 0

2 years ago

in rutherford's gold foil experiment a very small number of alpha particles were deflected. what about the structure of the atom

zvonat [6]

Answer:The atom being mostly empty space. A small number of alpha particles were deflected by large angles (> 4°) as they passed through the foil. There is a concentration of positive charge in the atom. Like charges repel, so the positive alpha particles were being repelled by positive charge

8 0

3 years ago

Put the substance in a 25 mL beaker.

kkurt [141]

Answer: 1. Liquid

2. oily

3. no (not a solid)

Explanation:

5 0

3 years ago

Read 2 more answers

Unscramble dreor to get a science answer

lesantik [10]

Order.
Hope this helps! :)

5 0

2 years ago

Read 2 more answers

The powder mixture (Cu, Al and Fe) = 10g was oxidized from sufficient chloride acid. 1) What are the possible reactions to the p

Mandarinka [93]

Answer:

1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

Explanation:

1) Possible reactions

2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Mass of each metal

a) Mass of Cu

The waste was the unreacted copper.

Mass of Cu = 2.5 g

b) Masses of Al and Fe

We have two relations :

Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g

H₂ from Al + H₂ from Fe = 6.38 L at NTP

i) Calculate the moles of H₂

NTP is 20 °C and 1 atm.

$\begin{array}{rcl}pV & = & n RT\\\text{1 atm} \times \text{6.38 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{293.15 K}\\6.38 & = & 24.06n \text{ mol}^{-1} \\n & = & \dfrac{6.38}{24.06 \text{ mol}^{-1} }\\\\ & = & \text{0.2652 mol}\\\end{array}$

(ii) Solve the relationship

Let x = mass of Al. Then

7.5 - x = mass of Fe

Moles of Al = x/27

Moles of Fe = (7.5 - x)/56

Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x /18

Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56

∴ x/18 + (7.5 - x)/56 = 0.2652

56x + 18(7.5 - x) = 267.3

56x + 135 - 18x = 267.3

38x = 132.3

x = 3.5 g

Mass of Al = 3.5 g

Mass of Fe = 7.5 g - 3.5 g = 4.0 g

The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

4 0

3 years ago