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2 years ago

Electrospray ionization mass spectroscopy often results in analyte molecules fragmenting.

Physics

1 answer:

guajiro [1.7K]2 years ago

5 0

Electrospray ionization mass spectroscopy often results in analyte molecules fragmenting

FALSE

<u>Electrospray ionization mass spectroscopy</u>

It is a technique using electrospray to generate ions for mass spectrometry by applying high voltage to the liquid.

Electrospray ionization is a soft ionization technique.

It is used for production of gas phase ions.

The process of Electrospray ionization mass spectroscopy

1. dispersal of a fine spray charge droplets

2. solvent evaporation

3. ion ejection from highly charged droplets

To know more about Electrospray ionization mass spectroscopy

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The vector magnitudes F and r are always postive, so the sign o W is determined entirely by the angle e between the force and the displacement.Submit Figure 1 off 1 part C

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What would happen if you took a really strong metal box and you started teleporting in hundreds and hundreds of small cotton bal

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Ryan places 0.150 kg of boiling water in a thermos bottle. How many kgs of ice at –12.0 °C must Ryan add to the thermos so that

Greeley [361]

Answer:

The value is $m_i = 0.0234 \ kg$

Explanation:

Generally from the calorimetry principle we have that

$Heat \ loss\ by\ ice = heat\ gained\ by\ water\$

So here heat gained water is mathematically represented as i.e

$Q_w = m_w * c_w * (T_w - T )$

substituting 0.150 kg for $m_w$ , 4200 J/kg.°C for $c_w$ , 100°C for $T_w$ and 75°C for T

We have

$Q_w = 0.150 * 4200 * (100 - 75 )$

$Q_w =15750 \ J$

The Heat loss by the ice is mathematically represented as

$Q_i = Q_1 + Q_2 + Q_3$

Here $Q_1$ is the energy to move the ice to its melting point which is evaluated as

$Q_1 = m_i * c_i * ( T_o -T_i)$

Here $m_i$ is the mass of ice

$c_i$ is the specific heat of ice with value $2.05 * 10^3 J/kg.°C$

$T_o$ temperature of ice at melting point with value 0°C

$T_i$ is the temperature of ice with value -12°C

$Q_2$ is the energy to move the ice from its its melting point to liquid which is evaluated as

$Q_2 = m_i * L$

Here L is the Latent heat of melting of ice with value $334 * 10^3 J/kg$

$Q_3$ is the energy to move the ice from liquid to the equilibrium temperature which is evaluated as

$Q_1 = m_i * c_w * ( T -T_o)$

$Q_i = m_i [ c_i * ( T_o -T_i) + L + c_w * ( T -T_o) ]$

=> $Q_i = m_i [ 2.05 * 10^3 * ( 0 -(-12)) + 334 * 10^3 + 4200 * ( 75 - 0) ]$

From

$Heat \ loss\ by\ ice = heat\ gained\ by\ water\$

We have that

$m_i * 673600 =15750$

=> $m_i = \frac{15750}{673600}$

=> $m_i = 0.0234 \ kg$

8 0

4 years ago

A sample of polonium-210 was left for 414 days.

jenyasd209 [6]

Answer:

PO(1) = PO(0) / 2 1 refers to 1 half life of PO(0)

PO(2) = PO(1) / 2 = P(0) / 4 amount of PO left after 2 half-lives

PO(3) = PO(2) / 2 = PO(0) / 8 amount of PO left after 3 half-lives

414 da / 138 da = 3 3 half-lives pass in 414 da

PO(0) = 8 PO(3) = 8 * 1.45E-4 g = 1.16E-3 g = .00116 g after 414 days

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3 years ago

Does the magnetic force on the longer sides of the generator loop help or hinder the rotation of the generator loop?

Ilia_Sergeevich [38]

Answer:

Help

Explanation:

We can observe using Fleming's right hand rule that magnetic field line are perpendicular to longer arms and parallel to the shorter arms. Therefore, magnetic force on the long arm will help the rotation of the generator rather than hindering it. No force on the shorter arm as they are parallel to the field lines.

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