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Harrizon [31]
3 years ago
10

What is another name for a flask

Chemistry
2 answers:
amid [387]3 years ago
6 0

Answer:

Bottle, thermos, container, vessel

Explanation:

Semmy [17]3 years ago
5 0

Answer:

the another name for a flask is container

Explanation:

I hope it helped

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Please help me with this question number 5!!
shtirl [24]
The answer is B, the electrolysis of water. One cannot be the answer since the condensation of steam requires a loss of heat, an exothermic reaction. Numbers 2 and 3 cannot be correct either as they do not indicate energy in the chemical equation, and so it must be balanced throughout. Meaning that the only possible answer is B.
6 0
3 years ago
What is the net ion charge of beryllium?<br><br> +2<br> -2<br> +1<br> -1
miss Akunina [59]
+2, because Beryllium is in the Group II of the periodic table.

Hope this helps!
8 0
3 years ago
Read 2 more answers
A gas temperature of 14 C and a volume of 4.5L. If the temperature is raised ot 29 C and the pressure is not changed, what is th
zhenek [66]
T1 = 14⁰C = 273 +14= 287 K
T1 = 29⁰C =273 +29 = 302 K
V1 = 4.5 L
P = constant

T1/T2 = V1/V2
287/302 = 4.5/V2

V2= 302*4.5/287 = 4.7L
V2 = 4.7 L

7 0
3 years ago
What is the total charge of 20 electrons in zeka-Coulombs?
Naily [24]

Answer:

3.2\times 10^{-18}\ C

Explanation:

No of electrons, n = 20

We need to find the total charge. We can use quantization of electric charge in this case.

Net charge,

q = ne

e is charge on an electron

q=20\times 1.6\times 10^{-19}\\\\q=3.2\times 10^{-18}\ C

So, there are 3.2\times 10^{-18}\ C of charge in 20 electrons.

3 0
4 years ago
8g of aqueous Sodium Hydroxide reacts with 4g of aqueous Aluminum Chloride to produce aqueous Sodium Chloride and solid Aluminum
slava [35]

Answer:

The limiting reactant is AlCl₃ and the excess reactant is NaOH.

Explanation:

  • The balanced equation for the mentioned reaction is:

<em>3NaOH(aq) + AlCl₃(aq) → 3NaCl(aq) + Al(OH)₃(s)↓,</em>

It is clear that 3.0 moles of NaOH(aq) react with 1.0 mole of AlCl₃(aq) to produce 3.0 moles of NaCl(aq) and 1.0 mole of Al(OH)₃(s).

  • Firstly, we need to calculate the no. of moles of (8.0 g) of NaOH and (4.0 g) of AlCl₃:

no. of moles of NaOH = mass/molar mass = (8.0 g)/(40.0 g/mol) = 0.2 mol.

no. of moles of AlCl₃ = mass/molar mass = (4.0 g)/(133.34 g/mol) = 0.03 mol.

  • From stichiometry; NaOH reacts with AlCl₃ with (3: 1) molar ratio.

∴ 0.09 mol of NaOH (the remaining 1.1 mol is in excess) reacts completely with 0.03 mol of AlCl₃.

  • So,

<em>the limiting reactant is AlCl₃ and the excess reactant is NaOH.</em>

<em></em>

6 0
3 years ago
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