Which of the following body system does the immune system rely on to provide the first line of defense against pathogens

JOSE BOUGHT 750 BAGS OF PEANUTS FOR 375 HE INTENDS TO SELL EACH BAG FOR 0.15 MORE THAN HE PAID HOW MUCH SHOULD HE CHARGE FOR EAC

labwork [276]

He paid 2 dollars for each bag. Then add 15 cents for that and he would charge $2.15 for each bag.

7 0

2 years ago

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WILL GIVE BRAINLEST

Bad White [126]

Answer:

A

Explanation:

7 0

2 years ago

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What is the mass in grams of 2.50 mol of ammonia vapor, NH3?

Crank

Molar mass:-

$\\ \tt\longmapsto 14+3(1)=17g/mol$

Now

$\\ \tt\longmapsto Given\:mass=No\;of\:moles\times Molar\;mass$

$\\ \tt\longmapsto Given\:mass=17(2.5)$

$\\ \tt\longmapsto Given\:mass=42.5g$

7 0

2 years ago

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0

2 years ago

Reading the above graph to the nearest tenth of a milliliter, what is the volume of sodium hydroxide at the equivalence point?

Vlada [557]

Answer:

The volume of sodium hydroxide at the equivalence point is:

<u>14.9 mL of sodium hydroxide</u>.

Explanation:

<u>The equivalence point occurs when, in this case, the HCl is completely neutralized with the solution of NaOH, how you can see this doesn't occur in the last point but occurs in the nineteenth point, where the pH is no more acid (below to 7) but is 11 approximately</u>, then you must see in the X-axis from this point and you can see the volume is almost 15, by this reason I calculate the valor of 14.9 milliliters.

7 0

2 years ago