Will bromine react with sodium

Past due! Help please I choose the middle picture for my hypothesis being inclusive can you help explain why and how it shows it

MrMuchimi

Answer:

Explanation:

its supposed to be not supported not inclusive because you said mix

5 0

3 years ago

Suppose 9 glucose molecules enter glycolysis. Calculate the number of inorganic phosphate molecules required as well as the numb

Aleksandr-060686 [28]

Answer:

18 inorganic phosphates and 18 pyruvates.

Explanation:

Glycolysis is the pathway involved in the metabolism of sugar. It is enzyme catalyze and converts glucose into pyruvate and a Hydrogen ion.

The free energy released are the used to form high-energy molecules such as ATP and NADH.

The net ATP production of Glycolysis involving a molecule of Glucose are 2 ATPs ,2 NADH and 2 pyruvates.

If 9 molecules of glucose enter glycolysis,then it has to be multiplied by 9 to give 18 pyruvates and 18 net ATPs

6 0

3 years ago

Following the instructions in your lab manual, you have titrated a 25.00 mL sample of 0.0100 M KIO3 with a solution of Na2S2O3 o

vivado [14]

Answer:

For 1: The amount of potassium iodate that were titrated is $2.5\times 10^{-4}$ moles

For 2: The amount of sodium thiosulfate required is $1.25\times 10^{-4}$ moles

Explanation:

For 1:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$

Molarity of $KIO_3$ solution = 0.0100 M

Volume of solution = 25 mL

Putting values in above equation, we get:

$0.0100M=\frac{\text{Moles of }KIO_3\times 1000}{25}\\\\\text{Moles of }KIO_3=\frac{0.0100\times 25}{1000}=0.00025mol$

Hence, the amount of potassium iodate that were titrated is $2.5\times 10^{-4}$ moles

For 2:

The chemical equation for the reaction of potassium iodate and sodium thiosulfate follows:

$2KIO_3+Na_2S_2O_3\rightarrow K_2S_2O_3+2NaIO_3$

By Stoichiometry of the reaction:

2 moles of potassium iodate reacts with 1 mole of sodium thiosulfate

So, 0.00025 moles of potassium iodate will react with = $\frac{1}{2}\times 0.00025=0.000125mol$ of sodium thiosulfate

Hence, the amount of sodium thiosulfate required is $1.25\times 10^{-4}$ moles

6 0

3 years ago

Mixtures of benzene and cyclohexane exhibit ideal behavior. A solution was created containing 1.5 moles of liquid benzene and 2.

goblinko [34]

Answer:

Vapour pressure of cyclohexane at 50°C is 490torr

Vapour pressure of benzene at 50°C is 90torr

Explanation:

Using Raoult's law, pressure of a solution is defined by the sum of the product sbetween mole fraction of both solvents and pressure of pure solvents.

$P_{solution} = X_{A}P^0_{A}+X_{B}P^0_{B}$

In the first solution:

$X_{cyclohexane}=\frac{2.5mol}{2.5mol+1.5mol} =0.625$

$X_{benzene}=\frac{1.5mol}{2.5mol+1.5mol} =0.375$

$340torr = 0.625P^0_{A}+0.375P^0_{B}$ (1)

For the second equation:

$X_{cyclohexane}=\frac{3.5mol}{3.5mol+1.5mol} =0.700$

$X_{benzene}=\frac{1.5mol}{3.5mol+1.5mol} =0.300$

$370torr = 0.700P^0_{A}+0.300P^0_{B}$ (2)

Replacing (2) in (1):

$340torr = 0.625P^0_{A}+0.375(1233.3-2.333P^0_{A})$

$340torr = 0.625P^0_{A}+462.5-0.875P^0_{A}$

-122.5torr = -0.250P°A

$P^0_{A} = 490 torr$

Vapour pressure of cyclohexane at 50°C is 490torr

And for benzene:

$370torr = 0.700*490torr+0.300P^0_{B}$

$P^0_{B}=90torr$

Vapour pressure of benzene at 50°C is 90torr

3 0

3 years ago

Suppose that 0.250 mol of methane, CH4(g), is reacted with 0.400 mol of fluorine, F2(g), forming CF4(g) and HF(g) as sole produc

shtirl [24]

300 gallons

Explanation:

7 0

3 years ago