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Anuta_ua [19.1K]

3 years ago

12

Help please if link will report

1 answer:

kupik [55]3 years ago

4 0

All i know is the first one is electromagnet, the second one is attraction, the third is magnetism, the fourth is magnetize, the fifth is repulsion, the sixth is lines of force, the seventh is magnetic pole, the eighth is electromagnetism, the ninth is magnetic field, and the tenth is permanent magnet

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Hypernatremia due to excess water loss: A. Diabetes insipidus B. Dialysis fluid excess C. Hyperaldosteronism D. Older persons

alex41 [277]

Answer:

Hypernantremia due to excess water loss

A. Diabetes insipidus

A rare disorder that causes the body to make too much urine

#Brainly

-Levi

5 0

2 years ago

Read 2 more answers

Aqueous hydrochloric acid reacts with solid manganese(iv) oxide to form aqueous manganese(ii) chloride, liquid water and chlorin

maria [59]

4HCl + MnO₂ → MnCl₂ + 2H₂O + Cl₂

2Cl⁻ → Cl₂ + 2e⁻
MnO₂ + 4H⁺ + 2e⁻ → Mn²⁺ + 2H₂O

2Cl⁻ + MnO₂ + 4H⁺ → Cl₂ + Mn²⁺ + 2H₂O

7 0

4 years ago

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A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second l

madreJ [45]

Explanation:

From first source, kinetic energy ( $K.E_{1}$ ) ejected is 1 eV and wavelength of light is $\lambda$ .

From second source, kinetic energy ( $K.E_{2}$ ) ejected is 4 eV and wavelength of light is $\frac{\lambda}{2}$ .

Relation between work function, wavelength, and kinetic energy is as follows.

K.E = $\frac{hc}{\lambda} - \phi$

where, h = Plank's constant = $6.63 \times 10^{-34}$ J.s

c = speed of light = $3 \times 10^{8}$ m/s

Also, it is known that 1 eV = $1.6 \times 10^{-19}$ J

Therefore, substituting the values in the above formula as follows.

From first source,

$K.E_{1}$ = $\frac{hc}{\lambda} - \phi$

1 eV = $1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi$

$1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (1)

From second source,

$K.E_{2}$ = $\frac{hc}{\lambda} - \phi$

$4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi$

$6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (2)

Now, divide equation (2) by 2. Therefore, it will become

${6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}$

$3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}$ ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

$1.6 \times 10^{-19} = \frac{\phi}{2}$

$\phi$ = $3.2 \times 10^{-19}$

= 2 eV

Thus, we can conclude that work function of the metal is 2 eV.

3 0

4 years ago

Earths atmosphere is warmed by the green house effect because heat radiant from earths surface is _ by gasses in the atmosphere.

mars1129 [50]

I think it would be B) Adsorbed

4 0

3 years ago

Will NaCl be soluble or insoluble

attashe74 [19]

NaCl is salt to it is obviously SOLUBLE :)

6 0

4 years ago