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3 years ago

Create a list of 5 potential jobs that students of astronomy can obtain.

Chemistry

2 answers:

ozzi3 years ago

5 0

Answer:

Applications Engineer.

Business Development Manager.

Chief Systems Engineer.

Coating Technician.

Director of Business Development.

Tanzania [10]3 years ago

5 0

Applications Engineer.
Business Development Manager.
Chief Systems Engineer.
Coating Technician.
Director of Business Development.
Electrical Engineer.
Engineer, Modeling & Simulation.
EO/IR Systems Engineer.

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Need help Quickly Are volcanos and earthquakes both constructive and destructive forces

algol13

Answer:

yes

Explanation:islands and mountains

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3 years ago

What are some real world applications in which ph is an important factor?

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Medications for skin , and medication for babies

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3 years ago

A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre

givi [52]

Answer:

$\mathbf{0.02 M}$

Explanation:

$\text{So, from the given question:}$

$\text{EDTA will make complex with}$ $V^{+3}$ $\text{and the remaining EDTA will react with }$ $Ga^{+3}$

$\text{Hence, the total concentration of}$ $V^{+3}$ & $Ga^{+3}$ $\text{will be equivalent to EDTA concentration.}$

$V_{EDTA} = 25 \ mL$

$V_{V^{+3}} = 59.0 \ mL$

$V_{Ga^{+3}} = 13.0 \ mL$

$M_{EDTA} = 0.0680 \ M$

$M_{V^{+3}} = ???(unknown)$

$M_{Ga^{+3}} = 0.0400 \ M$

$V^{+3} + EDTA \to V[EDTA] + EDTA(Excess) \to^{CoA} \ Ga[EDTA] _{complex}$

$M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})$

$0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18$

$x = \dfrac{1.18}{59}$

$\mathbf{x =0.02 \ M }$

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3 years ago

List some good ideas for Science Fair for Sixth Graders

grigory [225]

You could test how high the soda shoots up when you drop mentos in different types of soda. i did it when i was in 6th grade

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4 years ago

2 NH3 + 3 CuO →3 Cu + N₂ + 3 H₂O

Tatiana [17]

40.1g of nitrogen gas is produced.

The equation given is

2 NH₃ + 3 CuO →3 Cu + N₂ + 3 H₂O

This equation is already balanced.

When 3 moles of CuO are consumed, 1 mole of nitrogen gas is produced.

We get 1 mole of nitrogen from 3 moles of copper oxide.

We need to find the number of moles of nitrogen gas produced when 4.3 moles of copper oxide are consumed.

4.3/3 x 1 = 1.433 mols

1.433 mols of nitrogen gas are produced
The molar mass of nitrogen gas is 14+14 = 28g
The amount of nitrogen gas produced in grams is 28x1.433 = 40.1g

40.1g of nitrogen gas can be made when 4.3 moles of CuO are consumed.

Learn more about molarity here:

brainly.com/question/24305514

#SPJ10

5 0

2 years ago