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1 year ago

How many moles of water are in 1.23 x 10 to the 18th power water molecules

Chemistry

2 answers:

Ksivusya [100]1 year ago

6 0

Answer:

The answer is 2.04x10^-6 moles of water.

Explanation:

Avogadro's number is defined as the number of particles found in an amount of substance per mole. It is the factor that relates the moles of a substance to the mass of that substance. We will use Avogadro's number to calculate the number of moles of water. as follows:

Avogadro's number = 1 mol = 6.022 x 10^23 particles.

we will use the conversion factor to calculate the number of moles of water:

Moles of water = 1.23x10^18 particles x (1mol/6.022x10^23 particles) = 2.04x10^-6 moles

mamaluj [8]1 year ago

3 0

Divide the number of molecules you have by, 6.022 x 10^23. This will give you the moles of water, or the moles of anything, since there is always 6.022 x 10^23 molecules in 1 mole of substance.

1.23x10^24 atoms/6.022x10^23 atom/mole = 2.04 mole H20

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The compound Xe(CF3)2 decomposes in a first-order reaction to elemental Xe with a half-life of 30.0 min. If you place 4.5 mg of

GenaCL600 [577]

Answer : The time passed by the sample is, $1.2\times 10^2\text{ min}$

Explanation :

Half-life = 30.0 min

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{30.0\text{ min}}$

$k=0.0231\text{ min}^{-1}$

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $0.0231\text{ min}^{-1}$

t = time passed by the sample = ?

a = initial amount of the reactant = 4.5 mg

a - x = amount left after decay process =0.25 mg

Now put all the given values in above equation, we get

$t=\frac{2.303}{0.0231}\log\frac{4.5}{0.25}$

$t=125.15\text{ min}=1.2\times 10^2\text{ min}$

Therefore, the time passed by the sample is, $1.2\times 10^2\text{ min}$

7 0

2 years ago

The shape of hair is determined in part by the pattern of disulfide bonds in keratin, its major protein. From first to last, arr

UNO [17]

Answer:

I.A thiol containing reagent and Heating(gentle)

II. Disulfide bond Breaker for Keratin

III. curling the hair physically

IV.A good oxidizing agent application

V. Disulfide bonds reformer for keratin to hold the shape

Explanation:

The hair is mostly made of Keratin which is a structural proteins with about 15% cysteine

In order to change the hair appearance on the outside a reducing agent is required containing a thiol as to reduce the formation of the sulfide and disulfide bridges formation in the hair so the that any changes can be applied without the body naturally turning it back, i-e allowing easier manipulation after this hairs are to be curled physically in order to keep this a oxidizing agent is applied and then finally a keratin reformer is added to hold the shape

4 0

1 year ago

Vinegar is a solution of acetic acid (the solute) in water (the solvent) with a solution density of 1010 g/L. If vinegar is 0.80

Damm [24]

Answer:

4.8 %

Explanation:

We are asked the concentration in % by mass, given the molarity of the solution and its density.

0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:

MW acetic acid = 60.0 g/mol

mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g

mass of solution = 1000 cm³ x 1.010 g/ cm³ (1l= 1000 cm³)

= 1010 g

% (by mass) = 48.00 g/ 1010 g x 100 = 4.8 %

6 0

1 year ago

Calculate the molarity of each solution.

weeeeeb [17]

Q1)
molarity is defined as the number of moles of solute in 1 L solution
the number of moles of LiNO₃ - 0.38 mol
volume of solution - 6.14 L
since molarity is number of moles in 1 L
number of moles in 6.14 L - 0.38 mol
therefore number of moles in 1 L - 0.38 mol / 6.14 L = 0.0619 mol/L
molarity of solution is 0.0619 M

Q2)
the mass of C₂H₆O in the solution is 72.8 g
molar mass of C₂H₆O is 46 g/mol
number of moles = mass present / molar mass of compound
the number of moles of C₂H₆O - 72.8 g / 46 g/mol
number of C₂H₆O moles - 1.58 mol
volume of solution - 2.34 L
number of moles in 2.34 L - 1.58 mol
therefore number of moles in 1 L - 1.58 mol / 2.34 L = 0.675 M
molarity of C₂H₆O is 0.675 M

Q3)

Mass of KI in solution - 12.87 x 10⁻³ g
molar mass - 166 g/mol
number of mole of KI = mass present / molar mass of KI
number of KI moles = 12.87 x 10⁻³ g / 166 g/mol = 0.0775 x 10⁻³ mol
volume of solution - 112.4 mL
number of moles of KI in 112.4 mL - 0.0775 x 10⁻³ mol
therefore number of moles in 1000 mL- 0.0775 x 10⁻³ mol / 112.4 mL x 1000 mL
molarity of KI - 6.90 x 10⁻⁴ M

8 0

1 year ago

(a) At what substrate concentration would an enzyme with a kcat of 30.0 s−1 and a Km of 0.0050 M operate at one-quarter of its m

Dmitrij [34]

The missing graph is in the attachment.

Answer: (a) [S] = 0.0016M

(b) Vmax = 3V; Vmax = $\frac{3V}{2}$ ; Vmax = $\frac{11V}{10}$

Explanation: Enzyme is a protein-based molecule that speed up the rate of a reaction. Enzyme Kinetics studies the reaction rates of it.

The relationship between substrate and rate of reaction is determined by the Michaelis-Menten Equation:

$V=\frac{V_{max}[S]}{K_{M}+[S]}$

in which:

V is initial velocity of reaction

Vmax is maximum rate of reaction when enzyme's active sites are saturated;

[S] is substrate concentration;

Km is measure of affinity between enzyme and its substrate;

(a) To determine concentration:

$0.25V_{max}=\frac{V_{max}[S]}{0.005+[S]}$

$0.25V_{max}(0.005+[S])=V_{max}[S]$

$0.00125+0.25[S]=[S]$

0.75[S] = 0.00125

[S] = 0.0016M

For a Km of 0.005M, substrate's concentration is 0.0016M.

(b) Still using Michaelis-Menten:

$V=\frac{V_{max}[S]}{K_{M}+[S]}$

Rearraging for Vmax:

$V_{max}=\frac{V(K_{M}+[S])}{[S]}$

(b-I) for [S] = 1/2Km

$V_{max}=\frac{V(K_{M}+0.5K_{M})}{0.5K_{M}}$

$V_{max}=\frac{V(1.5K_{M})}{0.5K_{M}}$

$V_{max}=$ 3V

(b-II) for [S] = 2Km

$V_{max}=\frac{V(K_{M}+2K_{M})}{2K_{M}}$

$V_{max}=\frac{V(3K_M)}{2K_M}$

$V_{max}=\frac{3V}{2}$

(b-III) for [S] = 10Km

$V_{max}=\frac{V(K_{M}+10K_M)}{10K_M}$

$V_{max}=\frac{V(11K_{M})}{10K_{M}}$

$V_{max}=\frac{11V}{10}$

(c) Being the affinity between enzyme and substrate, the lower Km is the less substrate is needed to reach half of maximum velocity.

Km of enzyme A is 2μM and of enzyme B is 0.5μM.

Enzyme B has lower Km than enzyme A, which means the first will need a lower concnetration of substrate to reach half of Vmax.

Analyzing each plot, notice that the red-coloured graph reaches half at a lower concentration, therefore, red-coloured plot is for enzyme B, while black-coloured plot is for enzyme A

3 0

1 year ago