Who discovered the structure of insulin?

If a chemist has 2.3 moles of arsenic (as) that has a molar mass of 74.92 g/mol, which would be the best way to find out how man

mr Goodwill [35]

The best way to determine the number of atoms of arsenic in the sample will be to multiply 2.3 by Avagadro's number.
This is because Avagadro's number is the number of particles one mole of any substance has, and its value is 6.02 x 10²³
If the number of moles of a substance are known, then multiplying by Avagadro's number will give the number of particles. In this case, this is 1.38 x 10²⁴.

3 0

3 years ago

What is the total amount of heat absorbed by 100.0

sergejj [24]

Answer:

Option (B) 6270J

Explanation:

The following were obtained from the question:

M = 100g

T1 = 30°C

T2 = 45°C

ΔT = 45 —30 = 15°C

C = 4.18J/g°C

Q=?

Q = MCΔT

Q = 100 x 4.18 x 15

Q = 6270J

Therefore, the total amount of heat absorbed is 6270J

7 0

3 years ago

What is the wavelength of radiation emitted when an electron goes from the n = 7 to the n = 4 level of the Bohr hydrogen atom? G

Phantasy [73]

Answer:

the wavelength of radiation emitted is $\mathbf{\lambda= 2169.62 \ nm}$

Explanation:

The energy of the Bohr's hydrogen atom can be expressed with the formula:

$\mathtt{E_n =- \dfrac{13.6\ ev}{n^2}}$

For n = 7:

$\mathtt{E_7 =- \dfrac{13.6\ ev}{7^2}}$

$\mathtt{E_7 =-0.27755 \ eV}$

For n = 4

$\mathtt{E_4=- \dfrac{13.6\ ev}{4^2}}$

$\mathtt{E_4 =- 0.85\ eV}$

The electron goes from the n = 7 to the n = 4, then :

$\mathtt{E_7-E_4 = (-0.27755 - (-0.85) ) \ eV}$

$\mathtt{= 0.57245\ eV}$

Wavelength of the radiation emitted:

$\mathtt{\lambda= \dfrac{hc}{0.57245 \ eV}}$

where;

hc = 1242 eV.nm

$\mathtt{\lambda= \dfrac{1242 \ eV.nm }{0.57245 \ eV}}$

$\mathbf{\lambda= 2169.62 \ nm}$

4 0

3 years ago

What element has the same number valence electrons as strontium?

Arturiano [62]

Answer:

Beryllium

Explanation:

They are in the same group

4 0

3 years ago

What is the mole fraction, X, of solute and the molality, m (or b), for an aqueous solution that is 15.0% NaOH by mass?

GalinKa [24]

Hello!

a) The mole fraction of solute of a 15% NaOH aqueous solution can be calculated in the following way:

First, we have to assume that we have 100 grams of solution. This will simplify the calculations.

Now, we know that this solution has 15 grams of NaOH and 85 grams of water. We can calculate the number of moles of each one in the following way:

$molesNaOH= 15 g NaOH*\frac{1molNaOH}{39,997 g NaOH}=0,3750 moles NaOH \\ \\ moles H_2O=85 gH_2O* \frac{1 mol H_2O}{18 g H_2O}=4,7222 moles H_2O$

To finish, we calculate the mole fraction by dividing the moles of NaOH between the total moles:

$X_{NaOH}= \frac{moles NaOH}{total moles}= \frac{0,3750 moles NaOH}{0,3750 moles NaOH+4,7222 molesH_2O} =0,073$

So, the mole fraction of NaOH is 0,073

b) The molality (moles NaOH/ kg of solvent) of a 15% NaOH aqueous solution can be calculated in the following way:

First, we have to assume that we have 100 grams of solution. This will simplify the calculations.

Now, we know that this solution has 15 grams of NaOH and 85 grams (0,085 kg) of water. We can calculate the moles of NaOH in the following way:

$molesNaOH= 15 g NaOH*\frac{1molNaOH}{39,997 g NaOH}=0,3750 moles NaOH$

Now, we apply the definition of molality to calculate the molality of the solution:

$mNaOH= \frac{moles NaOH}{kg_{solvent}}= \frac{0,3750 moles NaOH}{0,085 kg H_2O}=4,41 m$

So, the molality of this solution is 4,41 m

Have a nice day!

4 0

3 years ago