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earnstyle [38]
3 years ago
11

According to the cup wall diagrams, why does the double wall vacuum

Chemistry
1 answer:
sveta [45]3 years ago
8 0

Answer:

i do not now

Explanation:

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Help please and thanks
Likurg_2 [28]
The first one is 2
The second is 1
The third is 6
And the fourth is 3
4 0
3 years ago
Where does most of the mass of an atom come from?
Sedaia [141]

Where does most of the mass of the universe come from? In ordinary matter, most of the mass is contained in atoms, and the majority of the mass of an atom resides in the nucleus, made of protons and neutrons. Protons and neutrons are each made of three quarks.

7 0
3 years ago
CH4(g) + H2O(g) CO(g) + 3H2(g)
boyakko [2]
Presuming the arrow is between H20 and CO

On the left there are 2 gas moles.
On the right there are 4 gas moles.

The equilibrium will shift to the side with the most no. He gas moles when pressure is decreased.

Therefore the answer is A, since 4>2.

If you have any questions, feel free to ask
5 0
3 years ago
Water has a high specific heat because: Select one: a. it is a poor insulator b. hydrogen bonds must be broken to raise its temp
Crank
<h2>The required "option is b) hydrogen bonds must be broken to raise its temperature.</h2>

Explanation:

  • Water has high specific heat due to hydrogen bonds present in it.
  • The Ionisation of water does not affect the specific heat of the water.
  • On decreasing the temperature, there is the formation of bonds hence option (d) is wrong.
  • On increasing the temperature, there is the breaking of bonds hence option (b) is correct.
3 0
4 years ago
The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

K_a\times K_b=K_w

K_b for sodium formate is K_b=\frac {K_w}{K_a}

Given that:

K_a of formic acid = 1.8\times 10^{-4}

And, K_w=10^{-14}

So,

K_b=\frac {10^{-14}}{1.8\times 10^{-4}}

K_b=5.5556\times 10^{-11}

Concentration = 0.35 M

HCOONa    ⇒     Na⁺ +    HCOO⁻

Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

At t=0                              0.35                            -              -

At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

5.5556\times 10^{-11}=\frac {x^2}{0.35-x}

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

5 0
3 years ago
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