What may happen to the human body when exposed to an infectious agent? A. When infectious agents get into the human body, your body responds by functioning normally.

B. When infectious agents get into the human body, the body gets a surge of energy, causing a slight increase in body temperature, and you feel great.

C. When infectious agents get into the human body, your body responds by raising the core body temperature, causing a fever.

D.Nothing happens when the human body is exposed to an infectious agent.

The total mass of the solution of lithiumnitrate solution has been 99.7 grams.

Density can be defined as the mass of the solute per unit volume. The density can be expressed as g/ml or kg/L.

The mass of given Lithium nitrate = 2.5 grams.

The mass of water can be given as:

Density =

Volume of water = 97.2 ml.

The total mass of solution:

Mass of water = Density Volume

Mass of water = 1 97.2 grams

Mass of water = 97.2 grams

The total mass = Mass of lithium nitrate + mass of water

= 2.5 + 97.2 grams

= 99.7 grams.

The total mass of the solution of lithiumnitrate solution has been 99.7 grams.

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Answer:

The total mass of the solution = 99.7 g

Note: The question is incomplete. The complete question is given below:

A "coffee-cup" calorimetry experiment is run for the dissolution of 2.5 g of lithium nitrate placed into 97.2 mL of water. The temperature of the solution is initially at 23.5oC. After the reaction takes place, the temperature of the solution is 28.3 oC.  

1. Using a density of 1.0 g/mL for the water added and adding in the mass of the lithium nitrate, what is the total mass of the solution and solid?

Explanation:

mass = density * volume

density of water = 1.0 g/mL; volume of water = 97.2 mL

mass of water = 1.0 g/L * 97.2 mL

mass of water = 97.2 g

mass of lithium nitrate = 2.5 g

A solution is made by dissolving a solute (usually solid) in a solvent (usually a liquid). The solute in this reaction is lithium nitrate and the solvent is water.

Total mass of solution = mass of water + mass of lithium nitrate

Total mass of solution = 97.2 g + 2.5 g = 99.7 g

Therefore, total mass of the solution = 99.7 g

Answer:

ksp = 0,176

Explanation:

The borax (Na₂borate) in water is in equilibrium, thus:

Na₂borate(s) ⇄ borate²⁻(aq) + 2Na⁺(aq)

When you add just borax, the moles of Na²⁺ are twice the moles of borate²⁻, that means 2borate²⁻=Na⁺ (1)

The ksp is defined as:

ksp = [borate²⁻] [Na⁺]²

Then, borate²⁻(B₄O₇²⁻) reacts with HCl thus:

B₄O₇²⁻ + 2HCl + 5H₂O → 4H₃BO₃ + 2Cl⁻

The moles of HCl that reacts with B₄O₇²⁻ are:

0,500M×0,01200L = 6,00x10⁻³ mol of HCl

As two moles of HCl react with 1 mol of B₄O₇²⁻, the moles of B₄O₇²⁻ are:

6,00x10⁻³ mol of HCl× = 3,00x10⁻³ mol of B₄O₇²⁻

For (1), moles of Na⁺ are 3,00x10⁻³ mol ×2 = 6,00x10⁻³ mol of Na⁺

The [borate²⁻] is 3,00x10⁻³ mol of B₄O₇²⁻/0,00850L = 0,353M

And [Na⁺] is 6,00x10⁻³ mol of Na⁺ / 0,00850L = 0,706M

Replacing in the expression of ksp:

ksp = [0,353] [0,706]²

ksp = 0,176

I hope it helps!

Answer:

Option C

Explanation:

Avogadro number represent discrete unit of a substance which can be atom, molecule, ion etc.

In one mole of H2O there will be 6.02 x 1023 molecules and not atoms.

Molecule of H2O is the discrete unit of water mole and not atom.

Hence, option C is correct choice of answer

Answer: The lone pair of electron on nitrogen is accommodated in a 2p orbital hence it interacts with the pi system in aniline.

Explanation:

Aniline is less basic than amines. This is because, the nitrogen atom in aniline is not purely sp3 hybridized. Its actual hybrization state is closer to sp2 because the lone pair on nitrogen is accommodated in a 2p orbital. The nitrogen atim in aniline is planar and its

lonely pair interacts with the pi electron system of aniline. This makes the lone pair unavailable for protonation hence aniline is less basic than amines.

Final answer:

The calculated hybridization of the N-atom lone pair in aniline is affected by electron-electron repulsions, resonance, and steric effects from substituents on the aromatic ring.

Explanation:

The calculated hybridization of the N-atom lone pair in aniline is different from the predicted sp3 hybridization due to a combination of factors:

1. The presence of a lone pair on the nitrogen atom leads to electron-electron repulsions, causing distortions in the molecule's geometry.
2. The lone pair on the nitrogen atom can participate in resonance, resulting in delocalization of electrons and a change in hybridization.
3. The presence of the substituents on the aromatic ring can affect the hybridization of the N-atom lone pair by exerting steric effects.

Overall, these factors contribute to the observed hybridization of the N-atom lone pair in aniline, deviating from the predicted tetrahedral electron geometry.

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Answer:

x(t) = −39e

−0.03t + 40.

Explanation:

Let V (t) be the volume of solution (water and

nitric acid) measured in liters after t minutes. Let x(t) be the volume of nitric acid

measured in liters after t minutes, and let c(t) be the concentration (by volume) of

nitric acid in solution after t minutes.

The volume of solution V (t) doesn’t change over time since the inflow and outflow

of solution is equal. Thus V = 200 L. The concentration of nitric acid c(t) is

c(t) = x(t)

V (t)

=

x(t)

200

.

We model this problem as

dx

dt = I(t) − O(t),

where I(t) is the input rate of nitric acid and O(t) is the output rate of nitric acid,

both measured in liters of nitric acid per minute. The input rate is

I(t) = 6 Lsol.

1 min

·

20 Lnit.

100 Lsol.

=

120 Lnit.

100 min

= 1.2 Lnit./min.

The output rate is

O(t) = (6 Lsol./min)c(t) = 6 Lsol.

1 min

·

x(t) Lnit.

200 Lsol.

=

3x(t) Lnit.

100 min

= 0.03 x(t) Lnit./min.

The equation is then

dx

dt = 1.2 − 0.03x,

or

dx

dt + 0.03x = 1.2, (1)

which is a linear equation. The initial condition condition is found in the following

way:

c(0) = 0.5% = 5 Lnit.

1000 Lsol.

=

x(0) Lnit.

200 Lsol.

.

Thus x(0) = 1.

In Eq. (1) we let P(t) = 0.03 and Q(t) = 1.2. The integrating factor for Eq. (1) is

µ(t) = exp Z

P(t) dt

= exp

0.03 Z

dt

= e

0.03t

.

The solution is

x(t) = 1

µ(t)

Z

µ(t)Q(t) dt + C

= Ce−0.03t + 1.2e

−0.03t

Z

e

0.03t

dt

= Ce−0.03t +

1.2

0.03

e

−0.03t

e

0.03t

= Ce−0.03t +

1.2

0.03

= Ce−0.03t + 40.

The constant is found using x(t) = 1:

x(0) = Ce−0.03(0) + 40 = C + 40 = 1.

Thus C = −39, and the solution is

x(t) = −39e

−0.03t + 40.

The mass in grams of 3.011 x 10²³ atoms of F is 9.5 g.

The mass in grams of  1.50 x 10²³ atoms of Mg is 5.98 g.

The mass in grams of  4.50 x 10¹² atoms of Cl is 2.65 x 10⁻¹⁰ g.

The mass in grams of  8.42 x 10¹⁸ atoms of Br is 1.12 x 10⁻³ g.

The mass in grams of  25 atoms of W is 3.1 x 10⁻²¹ g.

The mass in grams of  1 atom of Au is 3.27 x 10⁻²² g.

What is the mass in grams of 3.011 x 10²³ atoms F?

The mass in grams of 3.011 x 10²³ atoms of F is calculated as follows;

6.023 x 10²³ atoms = 19 g of F

3.011 x 10²³ atoms F  = ?

= (3.011 x 10²³ x 19 g)/(6.023 x 10²³)

= 9.5 g

The mass in grams of  1.50 x 10²³ atoms of Mg is calculated as follows;

6.023 x 10²³ atoms = 24g of Mg

1.5 x 10²³ atoms F  = ?

= (1.5 x 10²³ x 24 g)/(6.023 x 10²³)

= 5.98 g

The mass in grams of  4.50 x 10¹² atoms of Cl is calculated as follows;

6.023 x 10²³ atoms = 35.5 g of Cl

4.5 x 10²³ atoms Cl  = ?

= (4.5 x 10¹² x 35.5 g)/(6.023 x 10²³)

= 2.65 x 10⁻¹⁰ g

The mass in grams of  8.42 x 10¹⁸ atoms of Br is calculated as follows;

6.023 x 10²³ atoms = 80 g of Br

8.42 x 10¹⁸ atoms Br = ?

= (8.42 x 10¹⁸  x 80 g)/(6.023 x 10²³)

= 1.12 x 10⁻³ g

The mass in grams of  25 atoms of W is calculated as follows;

6.023 x 10²³ atoms = 74 g of W

25 atoms W = ?

= (25  x 74 g)/(6.023 x 10²³)

= 3.1 x 10⁻²¹ g

The mass in grams of  1 atom of Au is calculated as follows;

6.023 x 10²³ atoms = 197 g of Au

1 atom of Au = ?

= (1  x 197 g)/(6.023 x 10²³)

= 3.27 x 10⁻²² g

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Final answer:

This solution provides the calculations necessary to convert the number of atoms of various elements (smallest particle of an element) to grams. It does so by using the molar mass of each element and Avogadro's number.

Explanation:

The mass of atoms can be determined by using Avogadro's number (6.022 x 1023 atoms/mol) and the molar mass of the specific element (g/mol). We use these to create a conversion factor and multiply by the number of atoms given.

1. For F (fluorine), which has a molar mass of about 18.9984 g/mol, 3.011 x 1023 atoms F is 9.00 g F.
2. For Mg (magnesium), with molar mass of about 24.3050 g/mol, 1.5 x 1023 atoms Mg is 6.07 g Mg.
3. For Cl (chlorine), with molar mass of about 35.453 g/mol, 4.50 x 1012 atoms Cl is 2.67 x 10-10 g Cl.
4. For Br (bromine), with molar mass about 79.904 g/mol, 8.42 x 1018 atoms Br is 0.12 g Br.
5. For W (tungsten), with molar mass about 183.84 g/mol, 25 atoms W is 7.65 x 10-22 g W.
6. For Au (gold), with molar mass about 197.0 g/mol, 1 atom Au is 3.28 x 10-22 g Au.

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