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viva [34]
1 year ago
12

What are 5 ways to prevent non communicable diseases?.

Chemistry
1 answer:
Semmy [17]1 year ago
3 0

Reduce risk factors that are easily altered,  as smoking, excessive alcohol consumption, poor diet, insufficient physical activity. Establishing ,enforcing sound legislative protections against non-communicable diseases .

<h3>What is a simple communicable disease?</h3>

A transmittable disease is one that can be spread from one generation to another via a number of routes, such as contact with blood or body fluids, inhalation of an airborne virus, or insect bites.

<h3>Why are contagious diseases significant?</h3>

The design and evaluation of illness preventive and control initiatives as well as the identification of common-source outbreaks depend on the reporting of communicable illness cases. Everybody will probably experience a communicable sickness at some time during their lives.

To know more about communicable diseases visit :

brainly.com/question/27330218

#SPJ4

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Choose all the answers that apply.
kipiarov [429]

Answer:

a. and b. both are correct. the number of electrons and neutrons IN THIS CASE of nitrogen Is same.

3 0
4 years ago
20 L of nitrogen gas are collected at a temperature of 50°C and 2 atm. How many grams of nitrogen gas were collected?
Andreas93 [3]

Answer:

0.1077 grams

Explanation:

First we will employ the ideal gas law to determine the number of moles of nitrogen gas.

PV=nRT

P=2 atm

V=20L

R=0.08206*L*atm*mol^-1*K^-1

T=323.15 K

Thus, 2atm*20L=n*0.08206*L*atm*mol^-1*K^-1*323.15K

K, atm, and L cancels out. Thus n=2*20mol/0.08206*323.15=1.5 moles

Lastly, we must convert the number of moles to grams. This can be done by dividing the number of moles by the molar mass of nitrogen gas, which is 14 grams.

1.5/14=0.1077 grams

8 0
3 years ago
Give chemical equation representing ionisation of carbonic acid
boyakko [2]

Answer:

HCO3− + OH− ⇌ CO32− + H2O

hope it helps!!!!

6 0
3 years ago
A 5 g sample of lead (specific heat 0.129 /g˚C) is heated, then put in a calorimeter with 50 mL of water (specific heat 4.184 J/
Svetach [21]

Answer:

670.68°C

Explanation:

Given that:

volume of water = 50 ml but 1 g = 1 ml. Therefore the mass of water (m) = 50 ml × 1 g / ml = 50 g

specific heat (C) = 4.184 J/g˚C

Initial temperature = 20°C, final temperature = 22°C. Therefore the temperature change ΔT = final temperature - initial temperature = 22 - 20 = 2°C

The quantity of heat (Q) used to raise the temperature of a body is given by the equation:

Q = mCΔT

Substituting values:

Q = 50 g × 4.184 J/g˚C × 2°C = 418.4 J

Since the mass of lead = 5 g and specific heat = 0.129 J/g˚C. The heat used to raise the temperature of water is the same heat used to raise the temperature of lead.

-Q = mCΔT

-418.4 J = 5 g × 0.129 J/g˚C × ΔT

ΔT = -418.4 J / ( 5 g × 0.129 J/g˚C) = -648 .68°C

temperature change ΔT = final temperature - initial temperature

- 648 .68°C = 22°C - Initial Temperature

Initial Temperature = 22 + 648.68 = 670.68°C

4 0
3 years ago
The number of proton, neutrons and electrons in oxygen atom
Ksenya-84 [330]

Answer:

in an oxygen atom there are:

protons:8

electrons:8

neutrons:8

Explanation:

this is because the atomic number of oxygen is 8 and that is the proton number and the electron number is the same as the atomic number

4 0
3 years ago
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