Write the balanced neutralization reaction that occurs between H2SO4

What element is being oxidized in the following redox reaction? Pb 2+(aq) + NH4 +(aq) - Pb(s) + NO3 (aq)

castortr0y [4]

The answer for your question is c.

5 0

3 years ago

The proportion of dissolved substances in seawater is usually expressed in

LiRa [457]

Answer:

The proportion of dissolved substances in seawater is usually expressed in ppm, ppb or ppt

Explanation:

The concentration of very diluted solutions should be expressed in parts per million, billion or trillion.

ppm = mass from the solute . 10⁶ / mass or volume of the solution

ppb = mass from the solute . 10⁹ / mass or volume of the solution

ppt = mass from the solute . 10¹² / mass or volume of the solution

ppm = mg/kg, μg/g, μg/mL → These are the units

ppb = ng/g

ppt = pg/g

3 0

4 years ago

Upon combustion, a 0.8009 g sample of a compound containing only carbon, hydrogen, and oxygen produces 1.6004 g CO2 and 0.6551 g

Ede4ka [16]

Answer:

C2H4O

Explanation:

We can get the answer through calculations as follows.

From the mass of carbon iv oxide produced, we can get the number of moles of carbon produced. We first divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles of carbon iv oxide is 1.6004/44 = 0.0364

Since there is only one carbon atom in CO2, the number of moles of carbon is same as above

The mass of carbon in the compound is simply the number of moles multiplied by the atomic mass unit. The atomic mass unit of carbon is 12. The mass of carbon in the compound is thus 12 * 0.0364= 0.4368g

From the number of moles of water, we can get the number of moles of hydrogen. To get the number of moles of water, we need to divide the mass of water by its molar mass. Its molar mass is 18g/mol. The number of moles here is thus 0.6551/18 = 0.0364 mole

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.0364= 0.0728

The mass of hydrogen is thus 0.0728* 1 = 0.0728g

The mass of oxygen equals the mass of the compound minus that of hydrogen and that of carbon.

= 0.8009 - 0.0728 - 0.4368 = 0.2913 mole

The number of moles of oxygen is the mass of oxygen divided by its atomic mass unit.

That equals 0.2913/16 = 0.0182 mole

The empirical formula can be obtained by dividing the number of moles of each by the smallest which is that of carbon and oxygen 0.0182

H = 0.0728/0.0182 = 4

C= 0.0364/0.0182 = 2

O= 0.0182/0.0182= 1

The empirical formula is thus C2H4O

3 0

3 years ago

2C2H2(g)+5O2(g)=4CO2(g)+2H2O(g)

balandron [24]

The volume of ethyne, C₂H₂ required to produce 12 moles of CO₂ assuming the reaction is at STP is 134.4 L

<h3>Balanced equation</h3>

2C₂H₂(g) + 5O₂(g) --> 4CO₂(g) + 2H₂O(g)

From the balanced equation above,

4 moles of CO₂ were produced by 2 moles of C₂H₂

<h3>How to determine the mole of C₂H₂ needed to produce 12 moles of CO₂</h3>

From the balanced equation above,

4 moles of CO₂ were produced by 2 moles of C₂H₂

Therefore,

12 moles of CO₂ will be produce by = (12 × 2) / 4 = 6 moles of C₂H₂

<h3>How to determine the volume (in L) of C₂H₂ needed at STP</h3>

At standard temperature and pressure (STP),

1 mole of C₂H₂ = 22.4 L

Therefore,

6 moles of C₂H₂ = 6 × 22.4

6 moles of C₂H₂ = 134.4 L

Thus, we can conclude that the volume of C₂H₂ needed for the reaction at STP is 134.4 L

Learn more about stoichiometry:

brainly.com/question/14735801

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3 0

2 years ago

The Michaelis constant for pancreatic lipase is 5 mM. At 60 C, lipase is subject to deactivation with a half-life of 8 min. Fat

Dmitriy789 [7]

Explanation:

According to the given data, we will calculate the following.

Half life of lipase $t_{1/2}$ = 8 min x 60 s/min

= 480 s

Rate constant for first order reaction is as follows.

$k_{d} = \frac{0.6932}{480}$

= $1.44 \times 10^{-3}s^{-1}
$

Initial fat concentration $S_{o}$ = 45 $mol/m^{3}$

= 45 mmol/L

Rate of hydrolysis $V_m_{o}$ = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration (S) = $S_{o} (1 - X)$

= 45 (1 - 0.80)

= 9 $mol/m^{3}$

or, = 9 mmol/L

It is given that $K_{m}$ = 5mmol/L

Therefore, time taken will be calculated as follows.

t = $-\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]$

Now, putting the given values into the above formula as follows.

t = $-\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]$

= $-\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s
}{K_{M} ln (\frac{45 mmol/L
}{9 mmol/L
}) + (45 mmol/L - 9 mmol/L
)]$

= $1642.83 s \times \frac{1 min}{60 sec}$

= 27.38 min

Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.

6 0

3 years ago