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seraphim [82]
3 years ago
6

How are the cells of the bicep muscle similar to the cells of the heart muscle?

Chemistry
1 answer:
Brums [2.3K]3 years ago
4 0

Answer:

I believe it's D

Explanation:

they both are worked a lot and it makes the most sense.

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The molar madd of Ca(OH)2
Nonamiya [84]
What you have to do is find a periodic table and add the mass of each atom that the compound is made of.

Ca= 40.1
O= 16.0
H= 1.01

keep in mind that you have to also account for how many atoms of each there are in the molecule. for example, in Ca(OH)2, there are one Ca, two O and two H

so the molar mass of Ca(OH)2= 40.1 + (2 x 16.0) + (2 x 1.01)= 74.12 g/mol
8 0
3 years ago
If the half life of iridium-182 is 15 years, how much of a 3 gram sample is left after 2 half-lives?
padilas [110]

Answer:

D. 0.75 grams

Explanation:

The data given on the iridium 182 are;

The half life of the iridium 182, t_{(1/2)} = 15 years

The mass of the sample of iridium, N₀ = 3 grams

The amount left, N(t) after two half lives is given as follows;

N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}} }

For two half lives, t = 2 × t_{(1/2)}

∴ t = 2 × 15 = 30

\dfrac{t}{t_{(1/2)}} = \dfrac{30}{15} = 2

\therefore N(t) = 3 \times\left (\dfrac{1}{2} \right )^2 = 0.75

∴ The amount left, N(t) = 0.75 grams

4 0
3 years ago
How many atoms are in 0.31 moles
padilas [110]
1 mole contains = 6.02x10^23 atoms. 0.31mole contains = 0.31x6.02x10^23 = 1.8662x10^23.
4 0
3 years ago
The three naturally occurring isotopes of potassium are 39K, 38.963707u; 40K, 39.963999u; and 41K.The percent natural abundances
sweet-ann [11.9K]

Answer:

The isotopic mass of 41K is 40.9574 amu

Explanation:

Step 1: Data given

The isotopes are:  

39K with an isotopic mass of 38.963707u and natural abundance of 93.2581%

40K with an isotopic mass of 39.963999u

41K wit natural abundance of 6.7302 %

Average atomic mass =39.098 amu  

Step 2: Calculate natural abundance of 40 K

100 % - 93.2581 % - 6.7302 %

100 % = 0.0117 %

Step 3: Calculate isotopic mass of 41K

39.098 = 38.963707 * 0.932581 + 39.963999 * 0.000117 + X * 0.067302

39.098 = 36.33681 + 0.0046758 +  X * 2.067302

X = 40.9574 amu

The isotopic mass of 41K is 40.9574 amu

8 0
4 years ago
What do genotype will appear in boxes 2 and 3?
VMariaS [17]
Tt is the genotype that will appear in boxes two and three.

If you look at the column and row that intersect to form boxes two and three, you will see that they are T and t. That is the best way I can describe it, sorry if it’s confusing.
3 0
3 years ago
Read 2 more answers
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