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3 years ago

You can speed up a chemical reaction by using a recyclable substance called areactant

2 answers:

8 0

The recyclable substance that speeds up chemical reactions is normally called catalyst or when it comes to reactions in living systems- an enzyme.

Vilka [71]3 years ago

5 0

The answer is True. You can speed up chemical reactions by using a recyclable substance called a reactant.

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The two structural members, one of which is in tension and the other in compression, exert the indicated forces on joint o. dete

melisa1 [442]

We're going to remedy it with the parallelogram law.
$\alpha$ = one hundred eighty - 30 - 70 = eighty degrees
R = sqrt(2^2 + 3^2 - 2(2)(three) cos(80)
R = three.30 kN, we can conclude now that the value of the ensuing of R is 3.30 kN

sin $\beta$ /3 = sin(eighty)/three.304 = 63.4 stages

3.3 kN
180 + 33.4 = 213.4 degrees
63.4 - 30 = 33.4

4 0

4 years ago

A 3-ft3 rigid tank initially contains saturated water vapor at 300°F. The tank is connected by a valve to a supply line that car

zhannawk [14.2K]

Explanation:

$Given\\V=3ft^3\\T_{1} =300^oF\\T_{L} =400^oF\\P_{L} =200psia\\V_{f} =0.5V$

We assume kinetic and potential energy changes are negligible and there is no work interactions.

a) Taking tank as a system, The energy balance can be define as

$E_{in}- E_{out}= E_{sys}$

$m_{i} h_{L} -Q_{out}=m_{2} u_{2}-m_{1} u_{1}$

The mass balance could be written as

$m_{in}- m_{out}= m_{sys} \\m_{i}= m_{2}- m_{1}$

The final pressure in the tank could be defined as following

$P_{2} =P_{sat300^oF}$

from standard steam table we know at

$T_{2}= T_{1}=300^oF\\ P_{2}=67.028psia$

From steam table at

$T_{1} =300^oF\\v_{1}= v_{g}=6.4663ft^3/lbm\\v_{1}= v_{g}=1099.8Btu/lbm\\ v_{f} =0.01745ft^3/lbm\\v_{f} =269.51Btu/lbm\\$

$at\\P_{L} =200psia\\T_{L}=400^oF\\h_{L}=1210.9Btu/lbm$

initial mass in the tank could be define as

$m_{1}=\frac{V}{v_{1} } \\m_{1} =\frac{3}{6.4663} =0.464lbm\\$

Final mass in the tank could be define as

$m_{2}=\frac{V_{f} }{v_{f} }+\frac{V_{g} }{v_{g} } \\ m_{2} =\frac{1.5}{0.01745} +\frac{1.5}{6.4663} =86.2lbm$

The amount of steam that has entered the tank

$m_{i}=m_{2}- m_{1}\\ m_{i}=86.2-0.464=85.74lbm$

The internal energy in final state could be defined as following

$U_{2}=m_{f} u_{f}+ m_{g} u_{g}\\ U_{2} =85.96*269.51+0.232*1099.8=23422Btu\\$

The heat transfer could be defined as following

$Q_{out}=m_{i} h_{L}+m_{1} u_{1}-m_{2} u_{2}\\ Q_{out}=85.74*1210.9+0.464*1099.8-23422=80910Btu$

6 0

3 years ago

Read 2 more answers

The blackbody curved for a star named Beta is shown below. What is the surface temperature of the star rounded to the nearest wh

marshall27 [118]

<span>Then, since the peak wavelength of the star Beta is 200nm, use Wein law and round 200 to the nearest WHOLE NUMBER. Hope that helps. </span>

3 0

4 years ago

Read 2 more answers

calculate the density of a neutron star with a radius 1.05 x10^4 m, assuming the mass is distributed uniformly. Treat the neutro

Orlov [11]

To develop this problem it is necessary to apply the concepts related to the proportion of a neutron star referring to the sun and density as a function of mass and volume.

Mathematically it can be expressed as

$\rho = \frac{m}{V}$

Where

m = Mass (Neutron at this case)

V = Volume

The mass of the neutron star is 1.4times to that of the mass of the sun

The volume of a sphere is determined by the equation

$V = \frac{4}{3}\pi R^3$

Replacing at the equation we have that

$\rho = \frac{1.4m_{sun}}{\frac{4}{3}\pi R^3}$

$\rho = \frac{1.4(1.989*10^{30})}{\frac{4}{3}\pi (1.05*10^4)^3}$

$\rho = 5.75*10^{17}kg/m^3$

Therefore the density of a neutron star is $5.75*10^{17}kg/m^3$

4 0

3 years ago

Your friend comes across a good deal to purchase a gold ring. She asks you for advice and for you to test the ring. The ring has

Luden [163]

Answer:

She is not getting a good deal.

Explanation:

The equation that relates heat with mass, specific heat and temperature change of an object is $Q=mc\Delta T$ .

Always convert temperature to Kelvin, although in our case it's not necessary because the $\Delta T$ will be the same, and we will leave the mass in grams because we will be getting $J/g^{\circ}C$ units for specific heat, which we can compare to the one given for gold.

We then calculate the specific heat of the object in question:

$c=\frac{Q}{m\Delta T}=\frac{94.8J}{(4.54g)(47.5^{\circ}C-23^{\circ}C)}=0.8523 J/g^{\circ}C$

Which shows it's not gold.

5 0

3 years ago