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N76 [4]
3 years ago
14

Which type of immunity provides lifetime protection for the body against a specific pathogen?

Chemistry
1 answer:
shutvik [7]3 years ago
7 0
Active immunity will be the answer
Good luck:)
You might be interested in
A graduated cylinder contains 20.8 mL of water. What is the new water level, in milliliters, after 35.2 g of silver metal is sub
stepladder [879]

<u>Answer:</u> The new water level of the cylinder is 24.16 mL

<u>Explanation:</u>

To calculate the volume of water displaced by silver, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of silver = 10.49 g/mL

Mass of silver = 35.2 g

Putting values in above equation, we get:

10.49g/mL=\frac{35.2g}{\text{Volume of silver}}\\\\\text{Volume of silver}=\frac{35.2g}{10.49g/mL}=3.36mL

We are given:

Volume of graduated cylinder = 20.8 mL

New water level of the cylinder = Volume of graduated cylinder + Volume of water displaced by silver

New water level of the cylinder = (20.8 + 3.36) mL = 24.16 mL

Hence, the new water level of the cylinder is 24.16 mL

5 0
3 years ago
URGENT!!! What is the balanced form of this equation?
Ostrovityanka [42]

Explanation: This is a reaction of oxidation of H_2O_2 in the presence of acidified KMnO_4. Acidified KMnO_4 is a strong oxidizing agent.

To balance out the H^+ on the reactant side, we write H_2O on the product side.

Balancing out the following reaction gives us:

2MnO_4^-+6H^++5H_2O_2\rightarrow 2Mn^{2+}+8H_2O+5O_2

5 0
3 years ago
Antarctica, almost completely covered in ice, has an area
Marysya12 [62]

<u>Answer:</u> The mass of ice is 2.39\times 10^{22}g

<u>Explanation:</u>

We are given:

Area of Antarctica = 5,500,000mi^2=5,500,000\times 2.59\times 10^{10}=142.45\times 10^{15}cm^2      (Conversion factor:  1mi^2=2.59\times 10^{10}cm^2  )

Height of Antarctica with ice = 7500 ft.

Height of Antarctica without ice = 1500 ft.

Height of ice = 7500 - 1500 = 6000 ft = 182.88\times 10^3cm     (Conversion factor:  1 ft = 30.48 cm)

To calculate mass of ice, we use the equation:

\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}

We are given:

Density of ice = 0.917g/cm^3

Volume of ice = Area × Height of ice = 142.45\times 10^{15}cm^2\times 182.88\times 10^3cm=26051.26\times 10^{18}cm^3

Putting values in above equation, we get:

0.917g/cm^3=\frac{\text{Mass of ice}}{26051.26\times 10^{18}cm^3}\\\\\text{Mass of ice}=(0.917g/cm^3\times 26051.26\times 10^{18}cm^3=2.39\times 10^{22}g

Hence, the mass of ice is 2.39\times 10^{22}g

5 0
3 years ago
When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.
sp2606 [1]

Answer:

Mass of CaCl₂ =  20 g

CaCO is presewnt in excess.

Mass of of CaCO₃ remain unreacted =  7.007 g

Explanation:

Given data:

Mass of calcium carbonate = 25 g

Mass of hydrochloric acid = 13.0 g

Mass of calcium chloride produced = ?

Chemical equation:

CaCO₃ + 2HCl  →  CaCl₂  + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

Number of moles of CaCO₃= 25.0 g / 100.1 g/mol

Number of moles of CaCO₃ = 0.25 mol

Number of moles of HCl:

Number of moles of  HCl = Mass /molar mass

Number of moles of HCl = 13.0 g / 36.5 g/mol

Number of moles of HCl = 0.36 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

                  CaCO₃         :               CaCl₂

                    1                 :               1

                 0.25              :            0.25

                HCl                :                CaCl₂

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.18 mol × 110.98 g/mol

Mass of CaCl₂ =  20 g

The calcium carbonate is present in excess.

                HCl                :                CaCO₃

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

So, 0.18 moles react with 0.36 moles of HCl.

The moles of CaCO₃ remain unreacted = 0.25 -0.18

The moles of CaCO₃ remain unreacted = 0.07 mol

Mass of of CaCO₃ remain unreacted = Moles × molar mass

Mass of of CaCO₃ remain unreacted = 0.07 mol × 100.1 g/mol

Mass of of CaCO₃ remain unreacted =  7.007 g

7 0
3 years ago
How much energy would be released if 450.0 g of the substance completely reacted with oxygen?
SOVA2 [1]

Answer:

1.1853 kJ

Explanation:

4 0
3 years ago
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