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Oliga [24]
4 years ago
13

Retry: 1.Examine Record A. Use the three basic rules to figure out the ages of the layers. In Chart A, list the layers from youn

gest to oldest with the youngest layer in the first row.

Chemistry
2 answers:
zubka84 [21]4 years ago
7 0
I, D, B, E, H, C, J, F, A, G, K
Harman [31]4 years ago
3 0

Answer:

C, E, H, A, B, F, K, D, L, G, J, I.

Explanation:

Hello,

In this case we are dating each layer based on its deepness as the deeper the layer is, the farther back in time it is (older). In such a way, the youngest layer is C and henceforth by going down, we find older and older layers no matter if the layer is horizontal or diagonal, we just go down by straight line, therefore, from youngest to oldest, the order turn out into:

C, E, H, A, B, F, K, D, L, G, J, I.

Best regards.

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Equation Given : Al^(3+) + Na3PO4 ==> 3Na^+ + AlPO4
Helga [31]

1 mols of Aluminium ion forms 1 mol aluminium phosphate

Molar mass of AlPO_4

  • 27+31+16(4)
  • 58+48
  • 106u

Moles of AlPO_4

  • 61µg/106
  • 0.000061/106
  • 5.75×10^{-7}
  • 57.5µmol

Moles of Al3+=57.5µmol

3 0
3 years ago
Br2(g) cl2(g)⇌2brcl(g) δh∘f for brcl(g) is 14. 6 kj/mol. Δs∘f for brcl(g) is 240. 0 j/mol
max2010maxim [7]

The Change in Gibb's free energy, ΔG for the reaction at 298K is; -56.92KJ.

<h3>Gibb's free energy of reactions</h3>

It follows from the Gibb's free energy formula as expressed in terms of Enthalpy and Entropy that;

  • ΔG = ΔH - TΔS

On this note, it follows that;

  • ΔG = 14.6 - (298× 0.24)

Hence, the Gibb's free energy for the reaction is;

  • ΔG = 14.6 - 71.52
  • ΔG = -56.92KJ

Remarks: The question requires that we determine the Gibb's free energy for the reaction at 298K.

Read more on Gibb's free energy;

brainly.com/question/13765848

5 0
2 years ago
Read 2 more answers
What is the molarity of a solution that contains 17 g of nh3 in 0.50 l of solution?
Sauron [17]

Answer:

  • 2.0 M

Explanation:

<u>1) Data:</u>

a) M = ?

b) mass of solue = 17 g

c) solute: NH₃

d) V = 0.5o liter

<u>2) Formulae:</u>

a) number of moles, n = mass in grams / molar mass

b) M = n / V (in liters)

<u>3) Solution</u>

a) Molar mass of NH₃ = 17.03 g/mol

b) n = mass in grams / molar mass = 17 g / 17.03 g/mol = 0.998 mol NH₃

c) M = n / V (in liters) = 0.998 mol / 0.50 liter = 1.996 M

d) Round to the appropiate number of significant figures, 2: 2.0 M.

Answer: 2.0 M

8 0
3 years ago
Divergent tectonic plate boundaries most commonly form one: A non-volcanic islands. B ocean trenches C continental mountains D o
Flauer [41]

Answer:

D. ocean ridges​

Explanation:

6 0
3 years ago
Read 2 more answers
(A tablet containing calcium carbonate and fillers with a mass of 1.631 g was dissolved in HCl. After the fillers were filtered
FrozenT [24]

Answer:

1) 0.825 grams is the mass of pure calcium carbonate product collected at the end of the experiment.

2) The mass % calcium carbonate in the tablet is 50.58%.

3) The mass % of calcium in calcium carbonate  is 40.00 %.

4) Mass of calcium in a tablet is 0.33 g.

Explanation:

Mass of tablet = 1.631 g

Mass of the watch glass = 46.719 g

Mass of the watch glass + mass of calcium carbonate = 47.544 g.

Mass of calcium carbonate = 47.544 g - 46.719 g = 0.825 g

0.825 grams is the mass of pure calcium carbonate product collected at the end of the experiment.

Mass of calcium carbonate = 0.825 g

Percentage of calcium in tablet:

\%=\frac{\text{Mass of calcium carbonate }}{\text{Mass of tablet}}\times 100

\%=\frac{0.825 g}{1.631 g}\times 100=50.58\%

The mass % calcium carbonate in the tablet is 50.58%.

Percentage of calcium in calcium carbonate :

Molar mass of calcium carbonate = 100 g/mol

Atomic mass of calcium = 40 g/mol

Percentage of calcium in calcium carbonate :

\%=\frac{\text{atomic mass of calcium}}{\text{Molar mass of}CaCO_3}\times 100

\%=\frac{40 g/mol}{100 g/mol}\times 100=40.00\%

The mass % of calcium in calcium carbonate  is 40.00 %.

Mass of calcium carbonate = 0.825 g

The mass % of calcium in calcium carbonate  is 40.00 %.

Mass of calcium in a tablet : x

40.00\%=\frac{\text{Mass of calcium in a tablet}}{0.825 g}\times 100

x=40.00\times \frac{0.825 g}{100}=0.33 g

Mass of calcium in a tablet is 0.33 g.

3 0
3 years ago
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