Please help me with dis

Which of the following is not a living thing A) carbon dioxide d) an amoeba c) a daisy d) a jellyfish

Artyom0805 [142]

A; Carbon Dioxide is not a living thing.

6 0

3 years ago

If the density of a metal bar is 3.47 g/mL and volume is 30 mL, calculate the mass of the metal bar in grams (g). Show the steps

Hitman42 [59]

Answer:

$m=104.1g$

Explanation:

Hello,

In this case, considering that the density is considered as the quotient of the mass and volume:

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

We can easily compute the mass by solving for it as follows:

$m=V*\rho =30 mL*3.47g/mL\\\\m=104.1g$

Best regards.

7 0

3 years ago

synergistic interlayer and defect engineering in vs2 nanosheets toward efficient electrocatalytic hydrogen evolution reaction

Sever21 [200]

It is stated that VS2 nanosheets with abundant defects and an extended (001) interlayer spacing as large as 1.00 nm, which is a 74% expansion as compared to that (0.575 nm) of the pristine counterpart, may be produced using a straightforward one-pot solvothermal process.

For the electrocatalytic hydrogen evolution reaction (HER), the interlayer expanded VS2 nanosheets display exceptional kinetic metrics, including a low overpotential of 43 mV at a geometric current density of 10 mA cm2, a small Tafel slope of 36 mV dec1, and long-term stability of 60 h without any current fading. The performance is even comparable to that of the commercial Pt/C electrocatalyst and significantly better than that of the pure VS2 with a typical interlayer gap.

Learn more about electrocatalytic here-

brainly.com/question/28463319

#SPJ4

4 0

2 years ago

A 27 kg iron block initially at 375 C is quenched in an insulated tank that contains 130kg of water at 26 C. Assume the water th

Bess [88]

Solution :

a). Applying the energy balance,

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

$0=\Delta U$

$0=(\Delta U)_{iron} + (\Delta U)_{water}$

$0=[mc(T_f-T_i)_{iron}] + [mc(T_f-T_i)_{water}]$

$0 = 27 \times 0.45 \times (T_f - 375) + 130 \times 4.18 \times (T_f-26)$

$t_f=33.63^\circ C$

b). The entropy change of iron.

$\Delta s_{iron} = mc \ln\left(\frac{T_f}{T_i} \right)$

$= 27 \times 0.45\ \ln\left(\frac{33.63 + 273}{375 + 273} \right)$

= -9.09 kJ-K

Entropy change of water :

$\Delta s_{water} = mc \ \ln\left(\frac{T_f}{T_i} \right)$

$= 130 \times 4.18\ \ln\left(\frac{33.63 + 273}{26 + 273} \right)$

= 10.76 kJ-K

So, the total entropy change during the process is :

$\Delta s_{tot} = \Delta s_{iron} + \Delta s_{water}$

= -9.09 + 10.76

= 1.67 kJ-K

c). Exergy of the combined system at initial state,

$X=(U-U_{0}) - T_0(S-S_0)+P_0(V-V_0)$

$X=mc (T-T_0) - T_0 \ mc \ \ln \left(\frac{T}{T_0} \right)+0$

$X=mc\left((T-T_0)-T_0 \ ln \left(\frac{T}{T_0} \right)\right)$

$X_{iron, i} = 27 \times 0.45\left(((375+273)-(12+273))-(12+273) \ln \frac{375+273}{12+273}\right)$

$X_{iron, i} =63.94 \ kJ$

$X_{water, i} = 130 \times 4.18\left(((26+273)-(12+273))-(12+273) \ln \frac{26+273}{12+273}\right)$

$X_{water, i} =-13.22 \ kJ$

Therefore, energy of the combined system at the initial state is

$X_{initial}=X_{iron,i} +X_{water, i}$

= 63.94 -13.22

= 50.72 kJ

Similarly, Exergy of the combined system at initial state,

$X=(U_f-U_{0}) - T_0(S_f-S_0)+P_0(V_f-V_0)$

$X=mc\left((T_f-T_0)-T_0 \ ln \left(\frac{T_f}{T_0} \right)\right)$

$X_{iron, f} = 27 \times 0.45\left(((33.63+273)-(12+273))-(12+273) \ln \frac{33.63+273}{12+273}\right)$

$X_{iron, f} = 216.39 \ kJ$

$X_{water, f} = 130 \times 4.18\left(((33.63+273)-(12+273))-(12+273) \ln \frac{33.63+273}{12+273}\right)$

$X_{water, f} =-9677.95\ kJ$

Thus, energy or the combined system at the final state is :

$X_{final}=X_{iron,f} +X_{water, f$

= 216.39 - 9677.95

= -9461.56 kJ

d). The wasted work

$X_{in} - X_{out}-X_{destroyed} = \Delta X_{sys}$

$0-X_{destroyed} =$

$X_{destroyed} = X_{initial} - X_{final}$

= 50.72 + 9461.56

= 9512.22 kJ

6 0

3 years ago

Convert 0.020 kg of Sn to mg of Sn

Lapatulllka [165]

hope that helps !

7 0

3 years ago