Answer:
okay let me know when i can help, ill be happy too help
Answer:
Average atomic mass = 85.557 amu.
Explanation:
Given data:
Percent abundance of Rb-85 = 72.15%
Percent abundance of Rb-87 = 27.85%
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (72.15×85)+(27.85×87) /100
Average atomic mass = 6132.75 + 2422.95 / 100
Average atomic mass = 8555.7 / 100
Average atomic mass = 85.557 amu.
Answer: An increase in the ratio of insulin to glucagon will increase the activity of --
- Acetyl-CoA carboxylase(+)
-Phosphofructokinase PFK2(+)
-Glycogen synthase(+)
- Hormone sensitive lipase (-). The hormone sensitive lipase activity is not increased with increased insulin activity.
Explanation: increased insulin - glucagon ratio is usually high in fed state.Insulin helps the cells absorb glucose, reducing blood sugar and providing the cells with glucose for energy. When blood sugar levels are too low, the pancreas releases glucagon. Glucagon instructs the liver to release stored glucose, which causes blood sugar to rise.
C. 4 decorated cupcakes.
Because 12 divided by 3 = 4, so you can have 4 cupcakes each with 3 cherries
There are 19.5 g Na in 71.4 g NaHCO₃
Calculate the <em>molecular mass of NaHCO₃</em>.
1 Na = 1 × 22.99 u = 22.99 u
1 H = 1 × 1.008 u = 1.008 u
1 C = 1 × 12.01 u = 12.01 u
3 O = 3 × 16.00 u = <u>48.00 u
</u>
TOTAL = 84.008 u
So, there are 22.99 g of Na in 84.008 g NaHCO₃.
∴ Mass of Na = 71.4 g NaHCO₃ × (22.99 g Na/84.008 g NaHCO₃) = 19.5 g Na