If points E and D are midpoints of triangle ABC, then the perimeter of the triangle is:AB + BC + CA = 48, or:
3 t + 3 t + 4 t + 4 t + 7 t + 6 = 4821 t = 48 - 621 t = 42t = 42 : 21t = 2Answer: A. 2.
Answer: The results of Rutherford's gold foil experiment gave him the evidence to arrive at two conclusions: (1) an atom was much more than just empty space and scattered electrons and (2) an atom consists of a positive charge at the center where most of its mass is placed.
Explanation:
Ernest Rutherford performed an experiment in which he passed alpha particles through a thin gold foil sheet. Through this sheet some of the particles passed on to the other side but some of them were reflected back.
This experiment was done by Rutherford to prove than an atom contains a tiny and heavy nucleus. He concluded that an atom has more than empty space and electrons were present in scattered form.
Also, he concluded that an atom consists of a positive charge at the center where most of its mass is placed.
Thus, we can conclude that the results of Rutherford's gold foil experiment gave him the evidence to arrive at two conclusions: (1) an atom was much more than just empty space and scattered electrons and (2) an atom consists of a positive charge at the center where most of its mass is placed.
Answer:
the formula is 164.088 g/mol
Explanation:
Answer:
1) The value of Kc:
C. remains the same.
2) The value of Qc:
A. is greater than Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium.
4) The concentration of N2 will:
B. decrease.
Explanation:
Hello,
In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:
1) The value of Kc:
C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:
2) The value of Qc:
A. is greater than Kc, since the reaction quotient is:
![Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=Qc%3D%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.
4) The concentration of N2 will:
B. decrease, since less reactant is forming the products.
Best regards.
