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3 years ago

The strongest forces of attraction occur between molecules of(1) HCl (3) HBr(2) HF (4) HI

Chemistry

1 answer:

Alexxandr [17]3 years ago

4 0

It would likely be HF. It displays hydrogen bonding.

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A student used 10 mL water instead of 30mL for the extraction of the NaCl from the mixture. How would this affect the calculated

Usimov [2.4K]

The amount of Nacl would be 3 times more

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3 years ago

Scientists use their to make observations

frosja888 [35]

Microscopes?
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3 years ago

Read 2 more answers

Conversion of minus 1 coulomb meter in debye

Viktor [21]

Answer:

-1 Coulomb meter = -2.997 × 10²⁹ Debye

Explanation:

Given:

Coulomb meter = -1 CM

Find:

In debye

Computation:

We know that,

1 Coulomb meter = 299,792,458,178,090,000,000,000,000,000 Debye

So,

-1 Coulomb meter = -299,792,458,178,090,000,000,000,000,000 Debye

-1 Coulomb meter = -2.997 × 10²⁹ Debye

3 0

2 years ago

Natural gas is stored in a spherical tank at a temperature of 13°C. At a given initial time, the pressure in the tank is 117 kPa

drek231 [11]

Answer:

1. the absolute pressure in the tank before filling = 217 kPa

2. the absolute pressure in the tank after filling = 312 kPa

3. the ratio of the mass after filling M2 to that before filling M1 = 1.44

The correct relation is option c ( $\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }$ )

Explanation:

To find -

1. What is the absolute pressure in the tank before filling?

2. What is the absolute pressure in the tank after filling?

3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?

As we know that ,

Absolute pressure = Atmospheric pressure + Gage pressure

So,

Before filling the tank :

Given - Atmospheric pressure = 100 kPa , Gage pressure = 117 kPa

⇒Absolute pressure ( p1 ) = 100 + 117 = 217 kPa

Now,

After filling the tank :

Given - Atmospheric pressure = 100 kPa , Gage pressure = 212 kPa

⇒Absolute pressure (p2) = 100 + 212= 312 kPa

Now,

As given, volume is the same before and after filling,

i.e. $V_{1}$ = $V_{2}$

As we know that, P ∝ M

⇒ $\frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }$

⇒ $\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }$

⇒ $\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378$ ≈ 1.44

Now, as we know that PV = nRT

As V is constant

⇒ P ∝ MT

⇒ $\frac{P}{T}$ ∝ M

⇒ $\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }$

So, The correct relation is c option.

6 0

2 years ago

To balance a chemical equation, it may be necessary to adjust the

lana66690 [7]

Answer:

A. Coefficients

Explanation:

that's the number in front of the molecules

7 0

3 years ago