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andriy [413]
3 years ago
7

Location of nucleus in skeletal smooth and cardiac

Chemistry
1 answer:
never [62]3 years ago
5 0
I don'know,but you should do somthing to solve your question.
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Mass box A = 10 grams; Mass box B = 5 grams; Mass box C—made of one A and one B How many boxes of B would be required to make 30
Iteru [2.4K]
Mass box C is 10+5. (So C is 15)

But if C was 30, how many times could you put B (5) into it?

30/5 = 6

You would need 6 boxes of B to make 30 grams of C.
5 0
3 years ago
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Information on caffeine molecules with 3 uses and 3 properties ​
Setler [38]

Answer:

it's uses

1)To improve mental alertness

2) For treating migraine headaches.

3) It increases the blood adrenaline levels

it's properties

1) It is white, odorless and hygroscopic crystalline solid

2) It tastes bitter

3) It is soluble on water

Don't forget to make me brainlist plz

4 0
3 years ago
What are the molarity and osmolarity of a 1-liter solution that contains half a mole of calcium chloride? How many molecules of
givi [52]

Answer:

Molarity = 0.5 M

Osmolarity = 0.5 x 2 = 1 Osmpl.

Molecules of Cl2 = 6.02 x 10^{23} / 4= 1.505 x 10^{23} no. of molecules

Explanation:

If we add half mole in 1L volume than molarity will obviously be 0.5 M.

The osmolarity is molarity multiplies by number of dissociates of solute that for CaCl2 are 2. So, 2 x 0.5 = 1

Half will be molecules of Ca and half will be of Cl2 for 0.5M.

6 0
4 years ago
How does ionization energy change as you
kykrilka [37]

Answer:

How does the energy required to remove an electron from an atom change as you move left to right in Period 4 from potassium through iron? ... A greater nuclear charge pulls the electrons closer to the nucleus, decreasing the atomic radius.

3 0
2 years ago
What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )
yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution :  Given,

Concentration (c) = 0.150 M

Acid dissociation constant = k_a=4.5\times 10^{-4}

The equilibrium reaction for dissociation of HNO_2 (weak acid) is,

                           HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+

initially conc.         c                       0         0

At eqm.              c(1-\alpha)                c\alpha        c\alpha

First we have to calculate the concentration of value of dissociation constant (\alpha ).

Formula used :

k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}

Now put all the given values in this formula ,we get the value of dissociation constant (\alpha ).

4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}

4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2

0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0

By solving the terms, we get

\alpha=0.0533

No we have to calculate the concentration of hydronium ion or hydrogen ion.

[H^+]=c\alpha=0.150\times 0.0533=0.007995 M

Now we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.007995 M)

pH=2.097\approx 2.1

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0
4 years ago
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