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2 years ago

A given sample of caffeine, C8H10N4O2, has 6.47 x 1022

Chemistry

1 answer:

Verdich [7]2 years ago

3 0

Answer:

8.08 × $10^{22}$ atoms of hydrogen

Explanation:

From the given data, we have two categories of variables (unknowns):

Amount (moles or mass) of Caffeine
Amount (moles or mass) of Hydrogen

The longer route would be to solve for these variables first, by determining the number of moles of the Caffeine sample (0.134 moles of C8H10N4O2) from the mass of carbon (12.89 g = 1.07 moles in 6.47 $10^{22}$ atoms) provided. And then solve for the number of atoms of H. [N.B: 1 mole of ANY substance = 6.022 × $10^{22}$ atoms ]

Alternatively and quicker, we can use the mole ratios of the Carbon:Hydrogen atoms in the compound.

8 : 10 ≡ 4 : 5

If 4 moles -- 6.47 $10^{22}$ atoms

Then 5 moles -- ?? atoms

⇒ No. of atoms of Hydrogen = $\frac{5*6.47*10^{22}}{4}$

=8.08 × $10^{22}$ atoms of hydrogen

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Titration reveals that 11.6 mL of 3.0M sulfuric acid are required to neutralize the sodium hydroxide in 25.00mL of NaOH solution

Deffense [45]

Answer:

Explanation:

Molarity of acid(volume of acid)(# of H ions)= molarity of base(volume of base)(# of OH ions)

M(v)(#)=M(v)(#)

sulfuric acid sodium hydroxide

H2SO4 NaOH

(3)(11.6)(2)=M(25)(1)

M=2.784

6 0

2 years ago

Which condition can cause excessive pressure on the high side of a self contained active recovery device

murzikaleks [220]

Answer:

What can cause excessive pressure on the high side of an active self-contained recovery device? A closed recovery tank inlet valve or excessive air or other non condensables in the recovery tank (either A or B) Portable refillable tanks or containers used to ship recovered refrigerants must meet what standard(s)?

Explanation:

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5 0

1 year ago

An irregularly shaped solid which has a mass of 10.283g was placed in a graduated cylinder containing an inert liquid. The initi

Gwar [14]

Liquid + Solid = 8.89 mL
V ( Solid ) = 8.89 mL - 6.26 mL = 2.63 mL
The density of the solid = m / V = 10.283 g / 2.63 mL =
= 3.9 g/mL = 3.9 g / cm³

3 0

2 years ago

Calculate the molarity of sodium chloride in a half-normal saline solution (0.45% NaCl). The molar mass of NaCl is

Rus_ich [418]

Answer:

0.077 M

Explanation:

Data Given :

The concentration of half normal (NaCl) saline = 0.45g / 100 g

So,

Volume of Solution = 100 g = 100 mL

Volume of Solution in Liter = 100 mL / 1000

Volume of Solution = 0.1 L

molar mass of NaCl = 58.44 g/mol

Molarity:

Molarity is the representation of the solution. It is amount of solute in moles per liter of solution and represented by M

Formula used for Molarity

M = moles of solute / Liter of solution . . . . . . . . . . (1)

Now to find number of moles of Nacl

no. of moles of NaCl = mass of NaCl / molar mass

no. of moles of NaCl = 0.45g / 58.44 g/mol

no. of moles of NaCl = 0.0077 g

Put values in the eq (1)

M = moles of solute / Liter of solution . . . . . . . . . . (1)

M = 0.0077 g / 0.1 L

M = 0.077 M

So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M

3 0

2 years ago

Explain the effects of nh3 and hcl on the cuso4 solution in terms of le chatelier's principle

Fittoniya [83]

The Principle of Le Chatelier states that if a system in equilibrium is subjected to a disturbance, the system will react in such a way that it will diminish the effect of that disturbance. Thus, when the concentration of one of the substances in an equilibrium system is changed, the equilibrium varies in such a way that it can compensate for this change.

For example, if the concentration of one of the reactants is increased, the equilibrium shifts to the right or to the side of the products. Also, if you add more reagents, the reaction will move even more to the right until the balance is re-established again, increasing the quantity of products.

In this way, adding HCl to a solution of CuSO4 will produce the following reaction:

CuSO4 (aq) + 2HCl (aq) ⇔ CuCl2 (aq) + H2SO4 (aq)

Initially the solution of CuSO4 in water will be blue, but when adding HCl the solution will change color to green, since the aqueous solutions of CuCl2 are green. By adding more HCl this color will intensify as the balance shifts to the right, producing more CuCl2 and H2SO4.

On the other hand, adding NH3 to a solution of CuSO4 will produce the following reaction

CuSO4 (aq) + 4NH3 (aq) ⇔ [Cu(NH3)4] SO4 (s)

Thus, by adding NH3 to the CuSO4 solution we will observe the formation of a precipitate corresponding to [Cu(NH3) 4] SO4. When adding more NH3, the formation of more precipitate will be observed as the equilibrium moves to the right, producing a greater quantity of [Cu (NH3) 4] SO4.

6 0

2 years ago