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k0ka [10]
3 years ago
8

HELP ME PLEASE!!!

Chemistry
1 answer:
krok68 [10]3 years ago
7 0

Answer:

Br (Bromine)

H (Hydrogen)

Explanation:

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Which statement correctly compares an atom of an alkali metal with an atom of the alkaline earth metal next to it on the periodi
Mumz [18]
<span>The alkali metal atom forms a +1 ion, while the alkaline earth metal atom forms a +2 ion. hope this helps</span>
6 0
3 years ago
A chemist who studies water samples did a demonstration of how to test for lead in water. She added a clear solution of potassiu
Lerok [7]

Answer:

The swirling yellow solid formed is lead iodide (PbI₂).

Explanation:

  • The reaction of potassium iodide (KI) with lead nitrate (Pb(NO₃)₂) will produce lead iodide (PbI₂) and potassium nitrate (KNO₃) according to the equation:

2KI + Pb(NO₃)₂ → PbI₂↓ + 2KNO₃

  • Lead iodide (PbI₂) is a yellow swirling precipitate that is formed from the reaction.
3 0
3 years ago
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he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

8 0
3 years ago
199.5 grams unrefined dark crystalline sugar to cups
Dmitry [639]

Answer:

0.9975 cup  

Step-by-step explanation:

"Unrefined dark crystalline sugar" is what non-chemists call "brown sugar."

200.0 g brown sugar = 1 cup

 199.5 g brown sugar = 199.5× 1/200 .0

 199.5 g brown sugar = 0.9975 cup

A standard measuring cup is not capable of this precision and, furthermore, the mass of brown sugar you can get into a cup depends on how tightly you pack it.

Your Mole Day cake will be fine if you use 1 cup of brown sugar as usual.

4 0
3 years ago
What volume is occupied by 0.104 mol of helium gas at a pressure of 0.91 atm and a temperature of 314 K ?
Aleksandr-060686 [28]

Answer:

The volume will be "2.95 L".

Explanation:

Given:

n = 0.104

p = 0.91 atm

T = 314 K

Now,

The Volume (V) will be:

= \frac{nRT}{P}

By putting the values, we get

= \frac{0.104\times 0.0821\times 314}{0.91}

= \frac{2.6810}{0.91}

= 2.95 \ L

7 0
2 years ago
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