To determine the length of each side,
employ the distance formula represented by the equation:
Distance = ((x2-x1)^2+(y2-y1)^2)^0.5
Calculating
<span>AB = 8 units BC = 6 units AC = 10 units
</span><span>MN =8units NO = 6 units MO = 10 units
</span><span>XY = 6.32 units YZ = 6.32 units XZ = 8.94 units
</span>JK = 4.47 units KL = 4.47 units JL = 6 units
1 The correct response is option b) triangles ABC and MNO are Congruent. <span>These triangles, ABC and MNO, have congruent side lengths.
</span>2 The answer is option c) rotation.
There is a rotation of 90º around the origin for triangles ABC and MNO, where B=N,
B=N
C----------O
A----------M
<span>The partial pressure of oxygen is 438.0 mmHg. The ideal gas equation is expressed as PV = nRT where P represents pressure, V denotes volume, n is the number of moles, R is the ideal gas constant (8.3144598 (L*kPa)/(K*mol)), and T signifies absolute temperature. To convert from Celsius to Kelvin, we have 43.4 + 273.15 = 316.55 K. For the pressure conversion from mmHg to kPa: 675.9 mmHg * 0.133322387415 = 90.11260165 kPa. When solving for n using the ideal gas equation, we derive n = PV / (RT) which provides n = 90.11260165 kPa * 16.2 L / (8.3144598 (L*kPa)/(K*mol) * 316.55 K)= 1459.824147 L*kPa / 2631.94225 (L*kPa)/(mol), resulting in n = 0.554656603 mol. Thus, we have 0.554656603 moles of gas particles. Next, we determine the contribution from oxygen. The atomic weight of oxygen is 15.999 g/mol, while argon is 39.948 g/mol, and the molar mass of O2 is 31.998 g/mol. We establish the relationships where M is the number of moles of O2, and 0.554656603 - M gives the number of moles of Ar. Setting up the equation: M * 31.998 + (0.554656603 - M) * 39.948 = 19.3, we solve for M resulting in 0.359424148 moles of oxygen out of 0.554656603 total moles. This leads to oxygen providing 0.359424148 / 0.554656603 = 0.648012024 or 64.8012024% of the total pressure of 675.9 mmHg. The partial pressure therefore calculates to 675.9 * 0.648012024 = 437.9913271 mmHg, rounded to 438.0 mmHg</span>
Explanation:
Elements provided:
F, Sr, P, Ca, O, Br, Rb, Sb, Li, S
Elements sharing similar reactivity belong to the same group in the periodic table, indicating that those in the same column exhibit comparable reactivity. Here are the identified groupings:
Li and Rb are alkali metals in group 1
Ca and Sr are alkaline earth metals in group 2
F and Br are halogens in group 7
O and S belong to group 6
P and Sb are classified in group 5 of the periodic table
Thus, these classifications illustrate elements with the same chemical characteristics.
