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Lesechka [4]
3 years ago
6

What are differences between vaccines and antibotics

Chemistry
2 answers:
Ilia_Sergeevich [38]3 years ago
8 0

Answer:

Basic differences between vaccines and antibiotics are presented in the explanation below.

Explanation:

Antibiotics are compounds used to treat bacterial, fungal and protozoan infections, whereas vaccines are those that play an important role in providing immunity to particular viral diseases, such as common cold or influenza. A vaccine can be a dead or inactivated virus, or a compound taken from them after purification. Moreover, vaccines are more likely to help prevent diseases before they occur. Meanwhile, antibiotics help treat those that have already occurred.

lara31 [8.8K]3 years ago
3 0

Answer:

Explanation:

Antibiotics are chemical agents that are used to treat bacterial infections. ... Vaccines are used to prevent infection, particularly viral infections.

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How many moles are in 3.57 x 10 32<br> particles Ca?
valentina_108 [34]

Answer:

\boxed {\boxed {\sf 592,826,303.6 \ moles \ of \ Ca }}

Explanation:

To convert from moles to particles, we must Avogadro's Number.

6.022*10^{23}

This number tells us the number of particles (atoms, ions, molecules, etc.) in 1 mole of a substance. In this case, it is the particles of calcium in 1 mole of calcium.

6.022*10^{23} \ particles \ Ca / mole \ Ca

We can use Avogadro's Number as a ratio or fraction.

\frac{6.022 *10^{23} \ particles \ Ca}{1 \ mol \ Ca}

Multiply this by the given number of particles (3.57*10³²).

3.57*10^{32} \ particles \ Ca*\frac{6.022 *10^{23} \ particles \ Ca}{1 \ mol \ Ca}

Flip the fraction so the particles of calcium will cancel out.

3.57*10^{32} \ particles \ Ca*\frac{1 \ mol \ Ca}{6.022 *10^{23} \ particles \ Ca}

3.57*10^{32} *\frac{1 \ mol \ Ca}{6.022 *10^{23} }

\frac{3.57*10^{32} \ mol \ Ca}{6.022 *10^{23} }

592826303.6 \ mol \ Ca

There are <u>592,826,303.6 moles</u> of calcium in 3.57*10³² particles of calcium.

8 0
3 years ago
For C2H5OH (liquid)So = 161 J/K.mol; for C2H5OH (gas)So = 282.6 J/K.mol. What is ΔS if 92 g of C2H5OH is evaporated?
ira [324]
Tye answer is 111.5 j/k
6 0
4 years ago
Pls help me I don’t understand it’s for my science class.
olchik [2.2K]

Answer:

gold - element

nickel- element

6 0
3 years ago
describe how 250 cm³ of 0.2 mol/dm³ H2SO4 could be prepared from 150 cm³ of 1.0mol/dm³ stock solution of the acid​
Reptile [31]

Answer:

Explanation:

250 cm^3 of 0.2 moldm-3 H2SO4 can be prepared from 150cm^3 of 1.0 moldm^-3 by dilution.

150cm^3 of the 1.0 moldm^-3 stock solution is measured out using a measuring cylinder and transferred into a 250 cm^3 standard volumetric flask and made up to mark. The resulting solution is now 250cm^3 of 0.2 moldm-3 H2SO4.

5 0
3 years ago
P is the pressure in atmospheres (atm), V is the volume in liters (L), n is the number of moles, R is the gas constant (0.0821 L
yKpoI14uk [10]

Answer:

The mass of the neon gas  m = 1.214 kg

Explanation:

Pressure = 3 atm = 304 k pa

Volume = 0.57 L = 0.00057 m^{3}

Temperature = 75 °c = 348 K

Universal gas constant = 0.0821 \frac{L . atm}{mol K}

We have to change the unit of this constant. it may be written as

Universal gas constant = 8.314 \frac{KJ}{mol K}

Gas constant for neon = \frac{8.314}{20} = 0.41 \frac{KJ}{kg K}

From ideal gas equation,

P V = m R T ------- (1)

We have all the variables except m. so we have to solve this equation for mass (m).

⇒ 304 × 10^{3} × 0.00057 = m × 0.41 × 348

⇒ 173.28 = 142.68 × m

⇒ m = 1.214 kg

This is the mass of the neon gas.

4 0
3 years ago
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