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sergij07 [2.7K]
3 years ago
13

What is the number assigned to an element on the Periodic Table?

Chemistry
1 answer:
spayn [35]3 years ago
6 0

Answer:

Element Atomic Number One number you will find on all periodic tables is the atomic number for each element.

Explanation:

<em>hope it helps :))</em>

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Identical wire loops are dipped into Liquid X and Liquid Y, so that a film of liquid forms across the loops (like the bubble sol
muminat

Answer:

a.

Explanation:

Assuming that Liquid X is considered to possess a greater viscosity as well as higher surface tension than liquid Y. Then, liquid X will tend to harbour more pressure inside the liquid.

In addition to that, the greater the surface tension, the greater the force required to expand the liquid's surface area.

This in turn makes the force required to make the loop 5% wider to be greater in FX rather than FY.

Thus, option a is the correct answer.

6 0
3 years ago
What was the scientific basis for the unit amu
Naya [18.7K]
Natural abundance of oxygen I think
4 0
3 years ago
Calcium is a group 2 element. Chlorine is a group 7 element. They form an ionic compound called
kondaur [170]

Answer:

CaCl2

Explanation:

For every calcium there's 2 chlorine

7 0
3 years ago
The thermite reaction reacts iron (III) oxide, Fe2O3, with aluminium powder,Al, the form aluminium oxide, Al2O3 and iron, Fe.
lara [203]

Answer:

<u><em>This answer assumes that the strated "16.0g of iron" was meant to be 16.0 grams of iron(III) oxide.</em></u>

Explanation:

To start, the thermite equation must be balanced.

I find:

1Fe2O3 + 2Al = 1Al2O3 + 2Fe

This tells us we need 2 moles of Al for every 1 mole of Fe2O3.  

Now calculate the moles of each reactant:

Moles Fe2O3:  16.0 g/159.7 g/mole = <u>0.100 moles Fe2O3</u>

Moles Al:  8.1 /26.98 g/mole = <u>0.300 moles Al</u>

The balanced equation says that in order to react all of the Fe2O3 we'd need twice that amount (in moles) of the Al.  (0.100 moles Fe2O3)*(2) = 0.200 moles Al.

<u>Which of the two reactants is the limiting reagent?</u>

We have more than enough moles of Al to react with 0.10 moles of Fe2O3.  (We have 0.300 moles Al and all we need is 0.200 moles to react with the 0.10 moles of Fe2O3.  <em>Fe2O3 is the limiting reagent.</em>

<u><em>Calculate the maximum mass of iron of iron that could be formed using these quantities of reactants.</em></u>

The balanced equation tells us that we will obtain 2 moles of Fe for every 1 mole of Fe2O3 consumed.  Since Fe2O3 is the limiting reagent, we will assume that it completely reacts.  That means 0.1 moles of Fe2O3 is reacted.  Since we expect twice that many moles of Fe, we should obtain 0.200 moles of Fe.  At 55.85 g/mole, we should obtain:

(0.200 moles Fe)*(55.85 g Fe/mole Fe) = 11.2 grams Fe

<em></em>

<em></em>

5 0
2 years ago
The temperature of 6.24 L of a gas is increased from 25.0°C to 55.0°C at constant pressure. The new volume of the gas is Questio
Sphinxa [80]

Answer:

Heating this gas to 55 °C will raise its volume to 6.87 liters.

Assumption: this gas is ideal.

Explanation:

By Charles's Law, under constant pressure the volume V of an ideal gas is proportional to its absolute temperature T (the one in degrees Kelvins.)

Alternatively, consider the ideal gas law:

\displaystyle V = \frac{n \cdot R}{P}\cdot T.

  • n is the number of moles of particles in this gas. n should be constant as long as the container does not leak.
  • R is the ideal gas constant.
  • P is the pressure on the gas. The question states that the pressure on this gas is constant.

Therefore the volume of the gas is proportional to its absolute temperature.

Either way,

V\propto T.

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1}.

For the gas in this question:

  • Initial volume: V_1 = \rm 6.24\; L.

Convert the two temperatures to degrees Kelvins:

  • Initial temperature: T_1 = \rm 25.0\;\textdegree{C} = (25.0 + {\rm 273.15})\; K = 298.15\;K.
  • Final temperature: T_1 = \rm 55.0\;\textdegree{C} = (55.0 + {\rm 273.15})\; K = 328.15\;K.

Apply Charles's Law:

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1} = \rm 6.24\;L \times \frac{328.15\; K}{298.15\;K} = 6.87\;L.

7 0
4 years ago
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