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3 years ago

6

The cells of smokers often use anaerobic respiration more than the cells of non-smokers. Suggest one chemical in cigarette smoke

that may cause this.

1 answer:

valentina_108 [34]3 years ago

8 0

Answer:

Nicotine

Explanation:

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What is hydrogen bonding

yKpoI14uk [10]

Answer:

a weak bond between two molecules resulting from an electrostatic attraction between a proton in one molecule and an electronegative atom in the other.

Explanation:

For example, in water molecules (H2O), hydrogen is covalently bonded to the more electronegative oxygen atom. Therefore, hydrogen bonding arises in water molecules due to the dipole-dipole interactions between the hydrogen atom of one water molecule and the oxygen atom of another H2O molecule.

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3 years ago

70 oz of protein how many grams of protein are

Snowcat [4.5K]

There is a bout 7 grams of protein in 1 ounce of meat, so 7*70=

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3 years ago

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If a sample of oxygen occupies a volume of 2.15 L at a pressure of 58.0 kPa and a temperature of 25°C, what volume would this sa

Anna35 [415]

The answer is A) 1.13 L

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4 years ago

How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and

Kay [80]

Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

$(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)$

The expression used will be:

$Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]$

where,

$Q$ = heat released for the reaction = ?

m = mass of benzene = 94.4 g

$c_{p,s}$ = specific heat of solid benzene = $1.51J/g^oC=1.51J/g.K$

$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC=1.73J/g.K$

$\Delta H_{fusion}$ = enthalpy change for fusion = $-9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g$

Now put all the given values in the above expression, we get:

$Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]$

$Q=-29427.312J=-29.4kJ$

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

3 0

3 years ago

How much energy (in Joules) is required to convert 129 grams of ice at −23.0 °C to liquid water at 18.0 °C?

Karo-lina-s [1.5K]

Answer:

The energy that is required for the process is:

6230.7 J + 42957 J + 9715.2 J = <u>58902.9 joules</u>

Explanation:

This is a calorimetry problem:

Q = m . C . ΔT

Q = heat; m = mas; C is the specific heat and

ΔT = Final T° - Initial T°

Q = C lat . m

Q = Heat

m = mass

C lar = Latent heat of fusion

First of all we calculate the heat for ice, before it takes the melting point. (from -23°C to 0°C)

Q = 129 g . 2.10 J/g°C . (0°C - (-23°C)

Q = 129 g . 2.10 J/g°C . 23°C → 6230.7 joules

Then, the ice has melted. To be melted and change the state it required:

Q = C lat . m

Q = 333 J/°C . 129 g → 42957 joules

And in the end, we have water that changed its T° from O°C to 18°C

Q = 129 g . 4.184 J/g °C . (18°C - 0°C)

Q = 9715.2 Joules

The energy that is required for the process is:

6230.7 J + 42957 J + 9715.2 J = 58902.9 joules

5 0

3 years ago