A student redid the experiment of mixing room-temperature water and warm

A typical human contains 5.00 l of blood, and it takes 1.00 min for all of it to pass through the heart when the person is resti

Mrrafil [7]

<span>(a) 0.0676 l (b) 67.6 cc So we've been told that 5.00 L of blood flows through the heart every minute and that the heart beats 74.0 times per minute. So that means that for every beat of the heart, 5.00 L / 74.0 = 0.067567568 L of blood flows through the heart. Rounding to 3 significant figures gives 0.0676 l. Converting from liters to cubic centimeters simply require a multiplication by 1000, so we have 67.6 cc of blood pumped per beat.</span>

5 0

2 years ago

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What are the approximate boiling points for the c2, c4, c6, and c8 alkanes?

lakkis [162]

Alkanes are hydrocarbons with general formula $C_nH_{2n+2}$ .

So alkane with two carbon atoms will be $C_2H_6$ , alkane with four carbon atoms will be $C_4H_{10}$ , alkane with six carbon atoms will be $C_6H_{14}$ , alkane with eight carbon atoms will be $C_8H_{18}$ .

As the number of carbon atoms increase, the surface area will increase and thus the vanderwaal forces will also increase, and the boiling point will also increase.

Thus the approximate boiling point order is: $C_8>C_6>C_4>C_2$

The approximate boiling points will be: 125° C >68° C > -1 °C > -89° C

5 0

2 years ago

A 6.5 l sample of nitrogen at 25◦c and 1.5 atm is allowed to expand to 13.0 l. the temperature remains constant. what is the fin

ollegr [7]

Since the temperature of the gas remains constant in the process, we can use Boyle's law, which states that for a gas transformation at constant temperature, the product between the gas pressure and its volume is constant:
$pV=k$
which can also be rewritten as
$p_1 V_1 = p_2 V_2$ (1)
where the labels 1 and 2 mark the initial and final conditions of the gas.

In our problem, $p_1 = 1.5 atm$ , $V_1 =6.5 L$ and $V_2 =13.0 L$ , so the final pressure of the gas can be found by re-arranging eq.(1):
$p_2 = p_1 \frac{V_1}{V_2}= (1.5 atm) \frac{6.5 L}{13.0 L}=0.75 atm$

Therefore the correct answer is
<span>1. 0.75 atm</span>

8 0

2 years ago

Xander's friend Corey has a skateboard that he rides at Newton's Skate Park.

DanielleElmas [232]

Answer:

0.5 m/s2

Explanation:

Step 1:

Data obtained from the question.

Total Mass = 60Kg

Net force = 30N

Acceleration =?

Step 2

Determination of the acceleration.

Force = Mass x Acceleration.

With the above equation, we can easily obtain the acceleration as follow:

30 = 60 x Acceleration

Divide both side by 60

Acceleration = 30/60

Acceleration = 0.5 m/s2

Now, we can thus say that the acceleration at that moment is 0.5 m/s2

8 0

2 years ago

Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit

snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity= (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

is

$Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s$

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

$Inlet volume = inlet velocity*area of circle=\pi r^{2}*inlet velocity$

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

$Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}$

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0

2 years ago