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3 years ago

What item would you use to pour medium amounts of liquid

1 answer:

6 0

PVA glue or elmers glue PVA Glue does all the basics fairly well, and has all the binding ability you'd expect from an adhesive.

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What is the mass of 2 moles of nitrogen gas, N2?

drek231 [11]

Answer:

28g

Solution. Hence, 2 moles of nitrogen atoms weigh 28g.

6 0

2 years ago

A catalyst works by(1) increasing the potential energy of the reactants(2) increasing the energy released during a reaction(3) d

valentina_108 [34]

The catalyst (4) decreases the activation energy required for a reaction, by holding reactants in place

7 0

2 years ago

Read 2 more answers

Change 0.00765 kL into mL

bagirrra123 [75]

Answer:

7650

Explanation:

formula- multiply the volume value by 1e+6

4 0

3 years ago

Read 2 more answers

Explain why there would be a need for lithium batteries?

Leviafan [203]

Answer:

The much higher power density offered by lithium ion batteries is a distinct advantage. Electric vehicles also need a battery technology that has a high energy density. ... Lithium ion cells is that their rate of self-discharge is much lower than that of other rechargeable cells such as Ni-Cad and NiMH forms.

Put this into your own words or teachers will make you redo it

6 0

2 years ago

Typical "hard" water contains about 2.0 x 10–3 mol of Ca2+ per liter. Calculate the maximum concentration of fluoride ion that c

malfutka [58]

Answer:

$[F^-]_{max}=4x10{-3}\frac{molF^-}{L}$

Explanation:

Hello,

In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

$Ca^{+2}(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)$

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:

$[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}} \\$

$[F^-]_{max}=4x10{-3}\frac{molF^-}{L}$

Best regards.

8 0

2 years ago