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Ronch [10]
2 years ago
12

Heather just drank 40.0 grams of water (H20). How many moles of water did she just drink? O a. 2.22 moles b. 45 moles O c. 40.0

moles O d. 0.45 moles O e.720 moles
Chemistry
1 answer:
Ksju [112]2 years ago
5 0

Answer:

This question begins with something, you should know: molar mass from water is aproximately 18 g/m, so if 18 grams of water are contained in 1 mole, the 40 grams occuped 2.22 moles. As you see, opcion a is the best!

Explanation:

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Assertion (A): An orbital cannot have more than two electrons, moreover, if an orbital has two electrons they must have opposite
VMariaS [17]

Answer: Its option "A" Both A and R are true and R is the correct explanation of A.

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7 0
3 years ago
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In class we derived the Gibbs energy of mixing for a binary mixture of perfect gases. We also discussed that the same result is
GaryK [48]

Answer:

Attached below

Explanation:

Free energy of mixing = ΔGmix = Gf - Gi

attached below is the required derivation of the

<u>a) Molar Gibbs energy of mixing</u>

ΔGmix = Gf - Gi

hence : ΔGmix = ∩RT ( X1 In X1 + X2 In X2 + X3 In X3 + ------- )

<u>b) molar excess Gibbs energy of mixing</u>

Ni = chemical potential of gas

fi = Fugacity

N°i = Chemical potential of gas when Fugacity = 1

ΔG = RT In ( a2 / a1 )  

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3 years ago
When an erythrocyte is removed from circulation the iron ion of hemoglobin?
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Transporting metals, ions, water-insoluble molecules, and hormones. .... When erythrocytes are removed from circulation,
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3 years ago
Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

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K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

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