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3 years ago

Describe how antibodies protect your body against disease when exposed to a virus

1 answer:

8 0

When children are small they are given vaccines that are usually dead viruses given to the body. These viruses don't cause damage to body and the body takes it as a real virus and prepare antibodies in the body but when a certain real disease or virus is in the body , the already presented antibodies fight with them for the protection of the body. These antibodies remain in the body so that when a disease or virus attacks the body the antibodies are already geared up to fight against them. Thus antibodies protect the body against invading microbes or viruses.

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There are a few different shapes of airplanes being considered in the design of a new airplane. One design would add protrusions

Ede4ka [16]

C. The design would not be cost-efficient; the plane would require more fuel due to increased thrust from higher drag.

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3 years ago

Explain why your hair sticks to a plastic (polyethylene) comb.

N76 [4]

Answer:

Due to the accumulation of static charges/due to static electricity

8 0

3 years ago

Read 2 more answers

The pH of a saturated solution of a metal hydroxide MOH is 10.15. Calculate the Ksp for this compound.

Airida [17]

Answer:

$Ksp=2.00x10^{-8}$

Explanation:

Hello!

In this case, since the pH of the given metal is 10.15, we can compute the pOH as shown below:

$pOH=14-pH=14-10.15=3.85$

Now, we compute the concentration of hydroxyl ions in solution:

$[OH^-]=10^{-pOH}=10^{-3.95}=1.41x10^{-4}M$

Now, since this hydroxide has the form MOH, we infer the concentration of OH- equals the concentration of M^+ at equilibrium, assuming the following ionization reaction:

$MOH(s)\rightarrow M^+(aq)+OH^-(aq)$

Whose equilibrium expression is:

$Ksp=[M^+][OH^-]$

Therefore, the Ksp for the saturated solution turns out:

$Ksp=1.41x10^{-4}*1.41x10^{-4}\\\\Ksp=2.00x10^{-8}$

Best regards!

4 0

2 years ago

Net Ionic equations

Goryan [66]

<span>Net Ionic equation: 3Zn^2+(aq) + 2PO4^3-(aq) ---> Zn3(PO4)2 (s)
hope it helps
</span>

4 0

3 years ago

What volume of 0.100 m sodium chloride must be added to 75.0 ml of 0.200 m lead(ii) nitrate to precipitate all of the lead ions?

Svetradugi [14.3K]

Reaction for precipitation is
Pb(NO3)2(aq) + 2Nacl(aq)→PbCL2(s)+2NaNO3(aq)
The given reaction is
The mol of Nacl required is =2×mol of Pb(NO3)2 which is present.
∴M(Nacl)×V(Nacl) =2×M(Pb(NO3)2 ×V(Pb(NO3)2
o.100M×V(Nacl)=2×0.200 MX75.0ML
V(Nacl = 300mL.

5 0

3 years ago