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Nikolay [14]
3 years ago
10

Passive prosthetics came about to meet what need?

Chemistry
1 answer:
ss7ja [257]3 years ago
4 0

Answer:

c.............. ......

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How do you calculate the atomic mass number ?
anastassius [24]

add up the mass of protons and neutrons

3 0
2 years ago
Read 2 more answers
Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction: 2NO2 → 2NO + O2 In a particular experiment at 300 °C, [
Setler [38]

Answer:

rate=-1.75x10^{-5}\frac{M}{s}

Explanation:

Hello,

In this case, for the given information, we can compute the rate of disappearance of NO₂ by using the following rate relationship:

rate=\frac{1}{2}*\frac{C_f-C_0}{t_f-t_0}

Whereas it is multiplied by the the inverse of the stoichiometric coefficient of NO₂ in the reaction that is 2. Moreover, the subscript <em>f</em> is referred to the final condition and the subscript <em>0</em> to the initial condition, thus, we obtain:

rate=\frac{1}{2}*\frac{0.00650M-0.0100M}{100s-0s}\\\\rate=-1.75x10^{-5}\frac{M}{s}

Clearly, it turns out negative since the concentration is diminishing due to its consumption.

Regards.

3 0
3 years ago
46. The number of orbitals in the sub-levels increases by even numbers.<br> a. TRUE<br> b. FALSE
frosja888 [35]

Answer:

It's true.

Explanation:

In sub-energy level : number of orbitals

  • s : 2
  • p : 6
  • d : 10
  • f : 14

.

6 0
2 years ago
2. How many nanoliters are in 2.87 x 10-10 gallons?
Morgarella [4.7K]

Answer:

1.09nL

Explanation:

Hello,

In this case, given that 1 gal equals 4 qt, 1 qt equals 0.9464 L and 1 L equals 1x10⁹ nL, the dimensional analysis turns out:

2.87x10^{-10}gal*\frac{4qt}{1gal} *\frac{0.9464L}{1qt}*\frac{1x10^9nL}{1L}\\  \\1.09nL

Best regards.

4 0
3 years ago
Suppose that on a dry, sunny day when the air temperature is near 37 ∘C,37 ∘C, a certain swimming pool would increase in tempera
ioda

Answer:

The fraction of water body necessary to keep the temperature constant is 0,0051.

Explanation:

Heat:

Q=m*Ce*ΔT

Q= heat  (unknown)

m= mass  (unknown)

Ce= especific heat (1 cal/g*°C)

ΔT= variation of temperature  (2.75 °C)  

Latent heat:

ΔE=∝mΔHvap

ΔE= latent heat

m= mass  (unknown)

∝= mass fraction (unknown)

ΔHvap= enthalpy of vaporization (539.4 cal/g)

Since Q and E are equal, we can match both equations:

m*Ce*ΔT=∝*m*ΔHvap

Mass fraction is:

∝=\frac{Ce*ΔT}{ΔHvap}

∝=\frac{(1 cal/g*°C)*2.75°C}{539.4 cal/g}

∝=0,0051

7 0
2 years ago
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