An object has a density of 2.0g/cm3. If the object were cut in half what would be the density of one of the pieces?

True or False? <br><br> A day is about 12 hours long

7nadin3 [17]

Answer:

false it's 24 hours long.

7 0

2 years ago

A trial of this decomposition experiment, using different quantities of reactants than those listed in the question above produc

Paul [167]

Answer : The volume of $O_2(g)$ produced at standard conditions of temperature and pressure is 0.2422 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of $O_2$ gas = (740-22.4) torr = 717.6 torr

$P_2$ = final pressure of $O_2$ gas at STP= 760 torr

$V_1$ = initial volume of $O_2$ gas = 280 mL

$V_2$ = final volume of $O_2$ gas at STP = ?

$T_1$ = initial temperature of $O_2$ gas = $25^oC=273+25=298K$

$T_2$ = final temperature of $O_2$ gas = $0^oC=273+0=273K$

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

$\frac{717.6torr\times 280mL}{298K}=\frac{760torr\times V_2}{273K}$

$V_2=242.2mL=0.2422L$

Therefore, the volume of $O_2(g)$ produced at standard conditions of temperature and pressure is 0.2422 L

5 0

3 years ago

Explain the different between a molecule and a compound.

nevsk [136]

Answer:

-A molecule is the smallesr part is compound whereas a compound is the combination of two or more atoms in a fixed proportion by wiehgt.

- A glass of water is an example of compound but a small portion of water can be called molecule.

5 0

3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

3 years ago

Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol

prohojiy [21]

Hey there!:

From the given data ;

Reaction volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar = g/L * mol/g

[E]t = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence kcat of xyzase is 5.7*10¹⁰ s⁻¹

Hope that helps!

4 0

3 years ago