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Pavlova-9 [17]
3 years ago
14

Help pls pls pls I really need this answer

Chemistry
1 answer:
blagie [28]3 years ago
5 0

Answer:

I never did this but i think its d 37.5٪ lmk if i got it right pls and srry if i didnt

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Which of the following metals will react with water to produce a metal hydroxide and hydrogen gas? a. Mg b. Li c. Al d. Pb
Stells [14]

I believe the answer is A) Mg.

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A. what is the hybridization of the central atom in cse2?
elixir [45]

The orbital hybridization of the central carbon atom in CSe2 is sp.

In chemical bonding, atomic orbitals may be combined to form appropriate hybrid orbitals suitable for bonding. The orbitals that combine during hybridization must be close enough in energy.

In the compound Cse2, carbon is the central atom bonded to two selenium atoms. The carbon atom in CSe2 is sp hybridized.

Learn more about orbital hybridization:  brainly.com/question/1869903

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2 years ago
HELP ASAP!
sweet [91]

1. Meteorologist predict the weather by using tools. They use these tools to measure atmospheric conditions that occurred in the past and present, and they apply this information to create educated guesses about the future weather. The best we can do is observe past and present atmospheric patterns and data, and apply this information to what we think will happen in the future. Meteorologists use the scientific method on a daily – and even hourly – basis!

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8 0
3 years ago
How do you convert volume to moles at STP?
Vanyuwa [196]

The conversion of volume to moles at STP is 1 mole.

The ideal gas equation is given as :

P V = n R T

where,

P = pressure of the gas

V = volume of the gas

n = ?

R = constant = 0.823 atm L / mol K

T = temperature

At STP , the pressure is 1 atm and the temperature is 273.15 K, the volume At STP is 22.4 L.

moles , n = P V / R T

n = ( 1 × 22.4 ) / (0.0823 × 273.15)

n = 1 mole

Thus, at STP , the number of moles is 1 mol.

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6 0
1 year ago
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0
3 years ago
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