The answer from this questions is the letter
B. He should not use his opinions as evidence.
Answer: a. two substances present; two phases present : Heterogeneous mixture
b. two substances present; one phase present
: Homogeneous mixture
c. one substance present; one phase present
: pure substance.
d. one substance present; two phases present: Heterogeneous mixture
Explanation:
A pure substance is a substance which contains definite composition of only one type of component. Hence, it cannot be separated by physical means.
Mixture is a substance which contains two or more than two types of components and they can be separated by physical means as well.
Homogeneous mixtures: It is a mixture that has uniform composition throughout the solution and the particle size or shapes are not different. There is no physical boundary between the dispersed phase and dispersion medium.
Heterogeneous mixtures: It is a mixture that has non-uniform composition throughout the solution and the particle size or shapes are also different. There is a physical boundary between the dispersed phase and dispersion medium.
She will most likely observe that the temperature
does not change during melting because the heat absorbed is used to overcome
intermolecular forces rather than to increase the kinetic energy of the
particles if she measures the temperature of the water in the beaker.
Answer:
b. Beta emission, beta emission
Explanation:
A factor to consider when deciding whether a particular nuclide will undergo this or that type of radioactive decay is to consider its neutron:proton ratio (N/P).
Now let us look at the N/P ratio of each atom;
For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6
For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4
For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.
For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.
Answer:
463.0 g.
Explanation:
- We can use the following relation:
<em>n = mass/molar mass.</em>
where, n is the mass of copper(ii) fluoride (m = 4.56 mol),
mass of copper(ii) fluoride = ??? g.
molar mass of copper(ii) fluoride = 101.543 g/mol.
∴ mass of copper(ii) fluoride = (n)(molar mass) = (4.56 mol)(101.543 g/mol) = 463.0 g.
