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3 years ago

Why do things dissolve so well in water ?

Chemistry

1 answer:

Goshia [24]3 years ago

4 0

Answer:

Explanation:

Water is called the universal solvent. It is a polar molecule (105 degree angle between the H atoms) that gives it a + and a - side so to speak....which allows it to 'pull apart' substances....overcome their intra-molecular attractions to each other ...i.e. disssovle them

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The radioactive atom 61/27 co emits a beta particle. write an equation showing the decay

pickupchik [31]

Answer:

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{28}^{60}\text{Ni}$

Explanation:

The unbalanced nuclear equation is

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, ?$

Let's write the question mark as a nuclear symbol.

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{Z}^{A}\text{X}$

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.

Then

60 = 0 + A, so A = 60 - 0 = 60, and

27 = -1 + Z, so Z = 27 + 1 = 28

Your nuclear equation becomes

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{28}^{60}\text{X}$

Element 28 is nickel, so the balanced nuclear equation is

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{28}^{60}\text{Ni}$

7 0

3 years ago

How many L of 3.0 M H2SO4 solution can be prepared by using 100.0 mL OF 18 M H2SO4?

max2010maxim [7]

Answer: A volume of 600 mL of 3.0 M $H_{2}SO_{4}$ solution can be prepared by using 100.0 mL OF 18 M $H_{2}SO_{4}$ .

Explanation:

Given: $V_{1}$ = ?, $M_{1} = 3.0 M\\$

$V_{2} = 100.0 mL$ , $M_{2} = 18 M$

Formula used to calculate the volume is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\3.0 M \times V_{1} = 18 M \times 100.0 mL\\V_{1} = \frac{18 M \times 100.0 mL}{3.0M}\\= 600 mL$

Thus, we can conclude that a volume of 600 mL of 3.0 M $H_{2}SO_{4}$ solution can be prepared by using 100.0 mL OF 18 M $H_{2}SO_{4}$ .

6 0

3 years ago

37. How many grams does 4.00 moles of H20 weigh?

ivanzaharov [21]

Answer:

72 g

Explanation:

n=m/mr

m=n*mr

=4*(2+16)

+72 g

4 0

4 years ago

If 1.0 M HIM HI is placed into a closed container and the reaction is allowed to reach equilibrium at 25∘C∘C, what is the equili

sattari [20]

The given question is incomplete. The complete question is

If 1.0 M HI is placed into a closed container and the reaction is allowed to reach equilibrium at 25∘C∘C, what is the equilibrium concentration of H2 (g). Given the equilibrium constant is 62.

Answer: The equilibrium concentration of $H_2$ is 0.498 M