Answer: 20 mg Te-99 remains after 12 hours.
Explanation: N(t) = N(0)*(1/2)^(t/t1/2)
N(t) = (80 mg)*(0.5)^(12/6)
N(t) = 20 mg remains after 12 hours
Answer:
depends on how many you have...
Explanation:
Answer:
As the y-intercept increases, the graph of the line shifts up;
As the y-intercept decreases, the graph of the line shifts down
Explanation:
There are two ways to think about this problem. The first way would be the graphical approach:
- if we only change the y-intercept, this means we keep the same slope;
- y-axis is the vertical axis;
- if we change the point at which the line crosses the y-axis, we either shift it upward for a higher y-intercept or downward for a lower y-intercept.
Now, thinking algebraically, a line has the following equation in a general form:

The y-intercept is essentially obtained when x = 0, then:
y = b:
- if we increase b value, the y value increases, so the graph shifts upward;
- if we decrease b value, the y value decreases, so the graph shifts downward.
Answer:

Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to figure out the required net ionic equation by firstly writing out the complete molecular equation between aspirin and sodium acetate:

Whereas acetic acid and sodium acetylsalicylate are formed. Now, we write the complete ionic equation whereby sodium acetate and sodium acetylsalicylate are ionized because they are salts yet neither aspirin nor acetic acid are ionized as they are weak acids:

Finally, for the net ionic equation we cancel out the sodium spectator ions to obtain:

Regards!
The metalloids are on the right side of the periodic table B, Si, Ge, As, Sb, Te, and At. The nonmetals are also on the right side next to the metalloids, there should be a He at the top right of the periodic table and there should be one more nonmetal at the top left of the periodic table that is H. And from the metals they are all on the middle next to the metalloids, starting from Li, Be, Na, and Mg as so on all of those are metals.