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tatyana61 [14]
3 years ago
5

Designed to be a measure of the amount of energy released by an earthquake

Chemistry
1 answer:
11Alexandr11 [23.1K]3 years ago
5 0
The answer is the moment magnitude scale
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Please help on this question
natulia [17]
Send a more clearer picture...
But I will tell u the system of the Hadley cells---
METEOROLOGY
a large-scale atmospheric convection cell in which air rises at the equator and sinks at medium latitudes, typically about 30° north or south.
7 0
3 years ago
3 Na2O(aq) + 2 Al(NO3)3(aq) ----> _ Compound A + _ Compound B
Olin [163]
3Na2O(at) + 2Al(NO3)3(aq) —> 6NaNO3(aq) + Al2O3(s)

This is a double replacement reaction and NaNO3 is aqueous because Na is an alkali metal, plus nitrate is in the solution. Both of these are soluble. Al2O3 is not soluble because it does not contain any element that is soluble and is hence the precipitate.

Hope this helped!
7 0
3 years ago
A 25.00 mL sample of the ammonia solution
musickatia [10]

Answer:

1.634 molL-1

Explanation:

The mol ration between NH3 and HCl is 1 : 1

Using Ca Va / Cb Vb = Na / Nb   where a = acid and b = base

Na = 1

Nb = 1

Ca = 0.208 molL-1

Cb = ?

Va = 19.64 mL

Vb = 25.00mL

Solving for Cb

Cb = Ca Va / Vb

Cb = 0.208 * 19.64 / 25.0

Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)

Using the dilution equation;

C1V1 = C2V2

Initial Concentration, C1 = ?

Initial Volume, V1 = 25.00 mL

Final Volume, V2 =  250 mL

Final Concentration, C2 = 0.1634 molL-1

Solving for C1;

C1 = C2 * V2 / V1

C1 = 0.1634 * 250 / 25.00

C1 = 1.634 molL-1

3 0
3 years ago
3CuSO4
Alex73 [517]

Answer:

I think it 28.8

Explanation:

am not sure

5 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
3 years ago
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