Answer:
P total = P1 +P2
Explanation:
765 mm Hg = 18 mm Hg water + ? mm Hg nitrogen
747 mm Hg nitrogen
Answer:
a. 408g de NH3 son necesarios
b. 2.42x10²³ moléculas de agua se obtienen
Explanation:
a. Basados en la reacción, 2 moles de NH3 producen una mol de (NH2)2CO. Para resolver esta pregunta debemos convertir la masa de (NH2)2CO a moles. Estas moles a moles de NH3 y su respectiva masa:
<em>Moles (NH2)2CO -Masa molar: 60.06g/mol-</em>
720g * (1mol / 60.06g) = 11.99 moles (NH2)2CO
<em>Moles NH3: </em>
11.99 moles (NH2)2CO * (2mol NH3 / 1mol (NH2)2CO) = 23.98 moles NH3
<em>Masa NH3 -17.031g/mol-:</em>
23.98 moles NH3 * (17.031g / mol) = 408g de NH3 son necesarios
b. 1mol de CO2 produce 1mol de agua. Se debe convertir el volumen a moles usando PV = nRT. Estas moles = Moles de agua. 1mol = 6.022x10²³ moléculas:
Asumiendo STP:
PV = nRT; PV / RT = n
Donde P = 1atm a STP
V = 9L
R = 0.082atmL/molK
T = 273.15K a STP
1atm*9L / 0.082atmL/molK*273.15K = n
n = 0.402 moles CO2 = Moles H2O
<em>Moléculas:</em>
0.402 moles agua * (6.022x10²³ moléculas / 1mol) =
<h3>2.42x10²³ moléculas de agua se obtienen</h3>
Answer:
32.9%
Explanation:
<u>The mass of water </u>present in the original sample can be calculated by the substraction:
- m H₂O = sample mass before heating - sample mass after heating
- m H₂O = 8.24 g - 6.20 g = 2.04 g
The <u>percent by mass of water</u> can be calculated with the formula:
- Water Mass / Original Sample Mass * 100 %
* 100% = 32.9%
Answer ; The question is missing in some details, but here are he details ;
The two naturally occurring isotopes of bromine are
81Br (80.916 amu, 49.31%) and
79Br (78.918 amu, 50.69%).
The two naturally occurring isotopes of chlorine are
37Cl (36.966 amu, 24.23%) and
35Cl (34.969 amu, 75.77%).
Bromine and chlorine combine to form bromine monochloride, BrCl.
Explanation:
The detaile calculation is as shown in the attachment.
