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Angelina_Jolie [31]
3 years ago
5

Predict the effect of a mutation which limits the production of pigment in hair

Chemistry
1 answer:
pogonyaev3 years ago
8 0

Answer:

E

Explanation:

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A 603.0 g/hr stream of liquid methyl alcohol, also called methanol, (CH3OH) at 8.10 atm and 26.0°C was held at constant pressure
mafiozo [28]

Answer:

Q=\frac{1385270 J}{hr}

Explanation:

Methanol's molecular weight: M=32 g/mol

Methanol's heat of vaporization: \Delta H_{vap}=69.69 kJ/mol

Ideal gas heat capacity (Cp):

Cp=\frac{5}{2}*R where R is the gas constant

Cp=\frac{5}{2}*8.314 \frac{J}{mol*C}=20.78\frac{J}{mol*C}

The heat needed to vaporize and bring the gs to 210°C is:

Q=\frac{603 g/hr}{32 g/mol}*(69690\frac{J}{mol}+20.78\frac{J}{mol*C}*(210-26)C)

Q=\frac{1385270 J}{hr}

4 0
3 years ago
The pH of normal rainwater is
riadik2000 [5.3K]
The pH of normal rainwater is 5.6. It's slightly acidic
3 0
3 years ago
Read 2 more answers
100 points to the person who can answer these 4 questions
Vinil7 [7]

Answer:

....,................................

Explanation:

1= A

2=D

3=C

4=C

8 0
3 years ago
What is the percentage composition when 10g of magnesium combines with 4g of nitrogen?
Ivanshal [37]
The %  composition   when 10g of magnesium combine   with  4g of   nitrogen  is  71.43%   magnesium   and  28.57 %  nitrogen

               calculation

%
  composition  =  mass  of an element  / total mass  x100
mass  of magnesium = 10 g
mass of nitrogen  =  4g

calculate  the  total  mass  used

=  10g of  Magnesium  + 4 g of  nitrogen = 14 grams

%   composition for  magnesium  is therefore  =  10/14  x100 = 71.43 %

%  
composition  for  nitrogen  is therefore = 4 /14  x100  =   28.57 %
7 0
3 years ago
If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams
Westkost [7]

Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of KCl_3 = 2.0 moles

Molar mass of O_2 = 32 g/mole

Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2KClO_3\rightarrow 2KCl+3O_2

From the balanced reaction we conclude that

As, 2 mole of KClO_3 react to give 3 mole of O_2

So, 2.0 moles of KClO_3 react to give \frac{2.0}{2}\times 3=3.0 moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = 214.0^oC=273+214.0=487K

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

\rho = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}

P=1.78atm

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

7 0
3 years ago
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