All categories

2 years ago

Identify the alcohol.

Chemistry

1 answer:

Verizon [17]2 years ago

6 0

Answer:

give us something else to work with or this is just the best guess

You might be interested in

Radio waves travel at the speed of light which is 3.00 x 10^8. How many minutes does it take for a radio message to reach saturn

ryzh [129]

Answer:

43.89 min

Explanation:

Given that:-

The speed of light = $3.00\times 10^8\ m/s$

The distance = $7.9\times 10^8\ km$

The conversion of distance in km to distance into m is shown below as:-

1 km = 1000 m

So,

Distance = $7.9\times 10^8\times 1000\ m=7.9\times 10^{11}\ m$

The relation between speed distance and time is shown below as:-

$Speed=\frac{Distance}{Time}$

Thus,

$3.00\times 10^8=\frac{7.9\times 10^{11}}{Time}$

$300000000\times time=10^{11}\times \:7.9\ s$

Time = 2633.33 seconds

Also, 1 s = 1/60 min

So,

Time= $\frac{2633.33}{60}\ min=43.89\ min$

3 0

3 years ago

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r

rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

= 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

$K=Ae^{\frac{-E}{RT}}$

If $K_1=Ae^{\frac{-E_1}{RT}} -------equation 1$ &

$K_2=Ae^{\frac{-E_2}{RT}} -------equation 2$

$\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}$

$E_1= E_2-RT*In(\frac{K_1}{K_2})$

$E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})$

$E_1 = 28957.39292$ $J/mol$

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0

3 years ago

bottle contains 425 mL of insulin at a concentration of 50.0 mg/mL . What is the total mass of insulin in the bottle?

Ulleksa [173]

Hey there!

C = m
------
V

50.0 mg/mL = mass
---------------
425 mL

mass = 50.0 * 425

mass => 21250 mg

3 0

3 years ago

student has calibrated his/her calorimeter and finds the heat capacity to be 14.2 J/°C. S/he then determines the molar heat capa

nika2105 [10]

Answer:

24.03 J/mol.ºC

Explanation:

For a calorimeter, the heat lost must be equal to the heat gained from water plus the heat gained from calorimeter, which has the same initial temperature as the water.

-Qal = Qw + Qc (minus signal represents that the heat is lost)

-mal*Cal*ΔTal = mw*Cw*ΔTw + Cc*ΔTc

Where m is the mass, C is the specific heat, ΔT is the temperature variation, al is from aluminum. w from water and c from the calorimeter. Cw = 4.186 J/gºC

-25.5*Cal*(22.7 - 100) = 99.0*4.186*(22.7 - 18.6) + 14.2*(22.7 - 18.6)

1971.15Cal = 1699.10 + 58.22

1971.15Cal = 1757.32

Cal = 0.89 J/g.ºC

The molar mass of Al is 27 g/mol

Cal = 0.89 J/g.ºC * 27 g/mol

Cal = 24.03 J/mol.ºC

6 0

3 years ago

What does the number on the top left of the symbol show?

kkurt [141]

Answer:

Atomic mass

Explanation:

3 0

3 years ago

Read 2 more answers