What is a pentadihide?

Which statement is correct concerning hereditary information?

Papessa [141]

Answer:

the answer to this question is option A mark me as brainliest

mark me as brainliest

3 0

3 years ago

Elements within the same group of the periodic table behave similarly because they have the same number of what

Darina [25.2K]

Same number of ions
Yeeeeeeeeee

4 0

3 years ago

Read 2 more answers

Which statement best describes the movement of atoms in solid

ser-zykov [4K]

Atoms according to the kinetic molecular theory, are described in a solid to simply be vibrating in fixed positions, and not moving rapidly in the container. Because of this, they take a fixed volume and have fixed shape.

7 0

3 years ago

For the following hypothetical reaction:

olga55 [171]

The moles of B that will be needed to convert 2 moles of A into as many moles of C as possible is 6 moles

Explanation

3A +9B → 5C

The moles of B are calculated using the mole ratio.

That is; from the equation above the mole ratio of A:B is 3:9

If the moles of A required is 2 moles therefore the moles of B

= 2 x9/3= 6 moles

3 0

3 years ago

Two Carnot engines are operated in series with the exhaust (heat output) of the first engine being the input of the second engin

OlgaM077 [116]

Answer:

(a) 140 F

(b) The temperature rise at the point where the heat is dumped is 2.51 degC

Explanation:

(a) Considering T1 the temperature of input of the first engine, T2 the temperature of the exhaust of the first engine (and input of the second engine) and T3 the exhaust of the second engine, if both engines have the same efficiency we have:

$\eta=1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}$

The temperatures have to be expressed in Rankine (or Kelvin) degrees

$1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}\\\\\frac{T_1}{T2}=\frac{T_2}{T_3}\\\\(T_2)^{2} =T_1*T_3\\\\T_2=\sqrt{T_1*T_3} =\sqrt{(459.67+260)*(459.67+40)}= \sqrt{719.67*499.67}\\\\ T_2=599 \, R= (599-459.67) ^{\circ} F=140^{\circ} F$

(b) The Carnot efficiency of the cycle is

$\eta_{c}=1-Th/Ts=1-(273+20)/(273+315)=0.502$

If the efficiency of the plant is 60% of the Carnot efficiency, we have

$\eta=0.6*\eta_{c}=0.6*0.502=0.302$

The heat used in the plant can be calculated as

$Q_i=W/\eta=750MW/0.302=2483MW$

And the heat removed to the heat sink is

$Q_o=Qi-W=2483-750=1733MW$

If the flow of the river is 165 m3/s, the heat per volume in the sink is

$\frac{Q_o}{f} =\frac{1733 MJ/s}{165 m3/s}= 10.5MJ/m3$

Considering a heat capacity of water C=4.1796 kJ/(kg*K) and a density ρ of 1000 kg/m3, the temperature rise of the water is

$\Delta Q=C*\Delta T\\\Delta T=(1/C)*\Delta Q\\\Delta T=(\frac{1}{4.1796\frac{kJ}{kgK} } )*10,500\frac{kJ}{m3}*\frac{1m3}{1000kg}\\\Delta T= 2.51 ^{\circ}C$

8 0

3 years ago