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Gelneren [198K]
3 years ago
8

Which of the following inter-particle forces has the lowest potential energy?

Chemistry
1 answer:
Lisa [10]3 years ago
4 0

Answer:

b. hydrogen bonding.

Explanation:

Hope it helps.

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Water is flowing in a pipe with a mass flow rate of 100.0 lb/h (M water H,O=18.016). What is the (i) Molar flow rate of H20 (gmo
Arada [10]

Answer:

i) 0,7 molH20/s

ii)11,2 g O/s

iii)1,4 g H/s

Explanation:

i) To find the molar flow rate of water, we just convert the mass of water to moles of water using its molecular weight(g/mol) and changing to the proper units (lb to grames and hours to seconds):

100 \frac{lb}{h}*\frac{453,5g}{1 lb}*\frac{1molH20}{18,016g}*\frac{1h}{3600s}=0,7\frac{molH20}{s}

ii) Now we just consider the oxygen in the water stream (for 1 mole of water there is 1 mole of oxygen):

100\frac{lb}{h} *\frac{453,5g}{1lb}*\frac{1 molH20}{18,016g}*\frac{1molO}{1molH20}*\frac{16gr}{1molO}*\frac{1h}{3600s}=11,2\frac{gO}{s}

iii)Just considering the hydrogen in the stream (for 1 mole of water there is 2 moles of hydrogen):

100\frac{lb}{h} *\frac{453,5g}{1lb}*\frac{1 molH20}{18,016g}*\frac{2molH}{1molH20}*\frac{1gr}{1molH}*\frac{1h}{3600s}=1,4\frac{gH}{s}

3 0
3 years ago
Which group of animals would be served best by the following adaptations? Large ears to dissipate heat. Kidneys adapted to check
ale4655 [162]
A.<span>animals that live in deserts </span>
4 0
3 years ago
How much heat is needed to change the temperature of 5g of water from 20 oC to 37 oC?
Dafna1 [17]

Answer:

The answer to your question is Q = 355.64 J

Explanation:

Data

Heat = Q = ?

Temperature 1 = T1 = 20°C

Temperature 2 = T2 = 37°C

mass = m = 5 g

Specific heat = Cp = 4.184 J/g°C

Formula

Q = mCp(T2 - T1)

-Substitution

Q = (5)(4.184)(37 - 20)

-Simplification

Q = (5)(4.184)(17)

-Result

Q = 355.64 J

5 0
3 years ago
Read 2 more answers
Student prepares a 0.207 g sample of the nonvolatile solute caryophyllene by dissolving the substance in 1.00 g of chloroform (K
Sergeeva-Olga [200]

Answer: 204.2 g/mol

Explanation:

Elevation in boiling point is given by:

\Delta T_b=k_b\times m

\Delta T_b =  Elevation in boiling point =  3.68^0C/m

k_b = boiling point constant  = 3.63^0C/m

m = molality

Molality=\frac{n\times 1000}{W_s}

where,

n = moles of solute = \text {given mass}{\text {molar mass}}=\frac{0.207}{M}

 W_s = weight of solvent in g  = 1g

Molality =\frac{0.207\times 1000}{M\times 1g}

Putting in the values we get,

3.68=3.63\times \frac{0.207\times 1000}{M\times 1g}

M=204.2g/mol

The molar mass of caryophyllene is 204.2 g/mol.

5 0
3 years ago
Convert the following temperatures to Kelvin:
maxonik [38]

Explanation:

A. 100°C to Kelvins

T(K)=T(^oC)+273.15

T(K)=100(^oC)+273.15=373.15 K

B 600°R to Kelvins

(T)^oK=((T)^oR)\times 1.8

(T)^oK=600\times 1.8 K = 1080 K

C. 98°F to Kelvins

(T(K)-273.15)=(T(^oF)-32)\times \frac{5}{9}

(T(K))=(98(^oF)-32)\times \frac{5}{9}+273.15=309.81K

D. 77.4°F to degree Celsius

((T)^oC)=((T)^oF-32)\times \frac{5}{9}

(T)^oC =(77.4^oF-32)\times \frac{5}{9}=25.22^oC

E. 77.4 K to degree Celsius

T(^oC)=T(^K)-273.15

T(^oC)=77.4(K)-273.15=-195.75^oC

F. 77.4°R to degree Celsius

(T)^oC=((T)^oR-491.67)\times \frac{5}{9}

(T)^oC=((77.4)^oR-491.67)\times \frac{5}{9}=-230.15 ^oC

7 0
4 years ago
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