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Neko [114]
3 years ago
5

Which of the following pairs of elements could possibly be found in the same group on the periodic table? (2 points) A forms a 2

+ ion; B is in group 2. A is a noble gas; B forms a 1− ion. A has 3 valence electrons; B is a halogen in group 17. A forms a 2− ion; B is a metal in group 2.
Chemistry
1 answer:
harina [27]3 years ago
8 0
The first option: A forms 2+ ion and B is in group two: ex) Ca and Mg

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Liquid nitrogen has a density of 0.808 g/mL and boils at 77 K. Researchers often purchase liquid nitrogen in insulated 195-L tan
Reika [66]

Answer:

The nswer to the question is

The maximum fraction of the air in the room that could be displaced by the gaseous nitrogen is 0.548 or 54.8 %

Explanation:

To solve the question we note that

The density of the liquid nitrogen = 0.808g/mL and the volume is 195 L tank (vaporised)

Therefore since density = mass/volume we have

mass = Density × volume = 0.808 g/mL × 195 L × 1000 ml/L =157560 g

In gaseous form the liquid nitrogen density =1.15 g/L

That is density = mass/volume and volume = mass/density = 157560 g/(1.15g/L)  or

volume = 137008.69565 L

The dimension of the room = 10 m × 10 m × 2.5 m = 250 m³ and

1 m³ is equivalent to 1000 L, therefore 250 m³ = 250 m³  × 1000 L/m³ = 250000L

Therefore fraction of the volume occupied by the gaseous nitrogen =

137008.69565 L/250000 L = 0.548

Therefore the gaseous nitrogen occpies 54.8% of the room

7 0
3 years ago
What is the result of multiplying (2.5 × 1010) × (3.5 × 10-7)?
Kitty [74]
The result of multiplying that equation is 70,700.
4 0
3 years ago
CAN YOU PLEASE HELP ASAP​
Helga [31]
Your answer would be D you welcome !
7 0
3 years ago
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A sample of aluminum is placed in a 25 ml graduated cylinder containing 10 mL of water. The level of water rises to 18 mL. Alumi
anastassius [24]

To determine the mass of the sample, first find the volume difference after and before the aluminum was placed, the volume change is equal to the volume of the submerged object, in this case aluminum.

Then knowing volume of aluminum and the density of it, we can solve for the mass.

D = m/v

Dv = m

2.7 g/ml • 8 ml = 21.6 grams.

4 0
3 years ago
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Calculate the pressure of 2.50 Liters of a gas at 25.0oC if it has a volume of 4.50 Liters at
MariettaO [177]

Answer:

P_2 =0.51  atm

Explanation:

Given that:

Volume (V1) = 2.50 L

Temperature (T1) = 298 K

Volume (V2) = 4.50 L

at standard temperature and pressure;

Pressure (P1) = 1 atm

Temperature (T2) = 273 K

Pressure P2 = ??

Using combined gas law:

\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\ \\ \dfrac{1 *2.5}{298} = \dfrac{P_2*4.5}{273}

0.008389261745 \times 273 = 4.5P_2

P_2 =\dfrac{0.008389261745 \times 273 }{4.5}

P_2 =0.51 \ atm

4 0
2 years ago
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