<u>Answer:</u> The volume of the gas when the pressure and temperature has changed is 0.811 L
<u>Explanation:</u>
To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.
The equation follows:

where,
are the initial pressure, volume and temperature of the gas
are the final pressure, volume and temperature of the gas
We are given:
Conversion factor used:
1 L = 1000 mL
1 atm = 760 mmHg
![P_1=1.00atm=760mmHg\\V_1=790mL=0.790L\\T_1=37^oC=[37+273]K=310K\\P_2=695mmHg\\V_2=?L\\T_2=18^oC=[18+273]K=291K](https://tex.z-dn.net/?f=P_1%3D1.00atm%3D760mmHg%5C%5CV_1%3D790mL%3D0.790L%5C%5CT_1%3D37%5EoC%3D%5B37%2B273%5DK%3D310K%5C%5CP_2%3D695mmHg%5C%5CV_2%3D%3FL%5C%5CT_2%3D18%5EoC%3D%5B18%2B273%5DK%3D291K)
Putting values in above equation, we get:

Hence, the volume of the gas when the pressure and temperature has changed is 0.811 L
The mass ca(Oh)2 needed to produce 1.5 l of 0.25 solution is calculated as follows
find moles
=moles = molarity x volume
moles is therefore = 1.5 x 0.25 = 0.375 moles
mass = moles x molar mass
0.375mol x 74.09 g/mol = 27.78 grams
Answer: It is because tyrosine kinases and BTK have similar solubilities
Explanation:
In column chromatography, components of a mixture are seperated based on their relative solubilities in two non-mixing phases.
In essence, tyrosine kinases and BTK are present in the eluate due to their similar solubility rates that arise from the similar chemical structure both possess (otherwise it would be impossible for the inhibitor meant for Tyrosine kinase to bind and also inhibits BTK)
Thus, the similar solubilities of both groups is the reason they could elute out of the column without being adsorped.
Answer:

Explanation:
Hello!
In this case, since isotopes of the same element have the same number of protons and electrons but different atomic mass, we can compute the number of neutrons by subtracting the number of protons to the atomic mass of the isotope; thus, for Ga-69 and Ga-71 (rounded up to whole numbers), we obtain:

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Answer:
The molarity (M) of the following solutions are :
A. M = 0.88 M
B. M = 0.76 M
Explanation:
A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.
Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)
= 27 + 3(16 + 1)
= 27 + 3(17) = 27 + 51
= 78 g/mole
= 78 g/mole
Given mass= 19.2 g/mole


Moles = 0.246

Volume = 280 mL = 0.280 L

Molarity = 0.879 M
Molarity = 0.88 M
B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr
Molar mass of KBr = 119 g/mole
Given mass = 235.9 g

Moles = 1.98
Volume = 2.6 L


Molarity = 0.762 M
Molarity = 0.76 M