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hoa [83]
2 years ago
5

Write the Hafferent modes by which diseases spread ​

Chemistry
2 answers:
slava [35]2 years ago
8 0
Person to Person. When an infected person comes in contact with or exchanges body fluids with a non-infected person.
Droplet Transmission.
Spread by skin.
Spread through body fluids or blood.
Airborne Transmission.
Contaminated Objects.
Vector-Borne Diseases.
4.Food and Drinking Water.



Mark as brainliest
Vinvika [58]2 years ago
6 0

Answer:

  • Person to Person. When an infected person comes in contact with or exchanges body fluids with a non-infected person. ...
  • Droplet Transmission. ...
  • Spread by skin. ...
  • Spread through body fluids or blood. ...
  • Airborne Transmission. ...
  • Contaminated Objects. ...
  • Vector-Borne Diseases. ...
  • 4.Food and Drinking Water.

Explanation:

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a metal weighing 50.0g absorbs 20.0j of heat when its temperature increases by 150.0°c . what is the specific heat of the metal
nadezda [96]

Answer:

2.677 J/g°C

Explanation:

Quantity of heat required to raise the temperature of the metal,

Q = mc∆T

where m is the mass of the metal, c is the metal's specific heat capacity, and ∆T is the change in temperature that accompanies the heat transfer process

20.0J = 50.0g × c × 150°C

c = 20.0J/(50.0g × 150°C)

c = 2.677 J/g°C

7 0
3 years ago
Which element in each pair below has the higher first ionization energy?
amm1812

Answer:

a2

Explanation:

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6 0
3 years ago
HELP ASAP! will give 15 points
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4 years ago
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A substance has a freezing point of –38°C and a boiling point of 356°C. At what temperature would this substance be in its liqui
VLD [36.1K]

Answer:

X. 80°

Explanation:

if the substance is below -38 then it is in the solid phase (-100° and -50°) and if it is above 356° it is a gas (375°). Therefore anything in the middle is a liquid, 80°.

7 0
3 years ago
A sample of pure NO2 is heated to 337?C at which temperature it partially dissociates according to the equation2NO2(g)?2NO(g)+O2
tatyana61 [14]

Answer: Kc =5.915.10^{-3}

Explanation: Kc is an equilibrium constant of a chemical reaction and is dependent upon the reagents and products concentration. For this reaction, Kc is:  Kc = \frac{[NO]^{2} .[O2]}{[NO2]^{2} }

To determine each concentration, we have to find each relation:

From the reaction, we know that [NO] = 2[O2] (1)

From the Ideal Gas Law,

P·V=ntotal·R·T

\frac{P}{RT}=\frac{ntotal}{V}

\frac{P}{RT} = \frac{n(NO2)}{V} + \frac{n(NO)}{V} + \frac{n(O2)}{V}

[NO2]+{NO}+[O2] = \frac{0.745}{0.082.610} = 0.015 mol/L

As [NO] = 2[O2]

[NO2]+2[O2]+[O2]=0.015

[NO2] = 0.015 - 3[O2] (2)

To resolve this equation, we will turn towards density of the mixture:

ρ = \frac{m(NO2)+m(NO)+m(O2)}{V}

Substituting mass for molar mass and number of molar,

ρ = M(NO2)·\frac{n(NO2)}{V}+M(NO)·\frac{n(NO)}{V}+M(O2)·\frac{n(O2)}{V}

Knowing the molar mass of each molecule:

ρ = 46·[NO2]+30·[NO]+32·[O2] = 0.525 g/L (3)

Substituting (1), (2) and (3) we have:

46(0.015 - 3[O2])+30(2[O2])+32.[O2]=0.525

32[O2]+60[O2]-138[O2]=0.0525-0.69

-48[O2]= -0.6375

[O2]=13.3.10^{-3}M

Calculating for [NO]

[NO]=2[O2]

[NO]=2.13.3.10^{-3}

[NO]=26.6.10^{-3}M

And finding [NO2],

[NO2]=0.015 - 3[O2]

[NO2]=0.015 - 3.13.3.10^{-3}

[NO2]=39.885.10^{-3}M

So, to calculate Kc:

Kc = (26.6.10^{-3})²· 13.3.10^{-3} / (39.885.10^{-3})²

Kc= 5.915.10^{-3}

The Kc is 5.915.10^{-3}.

6 0
3 years ago
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