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S_A_V [24]
2 years ago
12

Pls help ASAP I’ll mark BRAINLIEST this is science btw.

Chemistry
1 answer:
Dmitrij [34]2 years ago
3 0

Answer:

It should be Magnetic fields can push and pull objects without touching

but I only say with touching. Maybe it's an error on you're teachers/Prof behalf.

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A particular solution of NaOH has hydronium concentration of 4x10-9 M. what is the solution’s pH?
bearhunter [10]

Answer:

The pH of the solution is 8, 40-

Explanation:

The pH indicates the acidity or basicity of a substance. PH values between 0 and less than 7 indicate acidic solutions, 7 neutral and greater than 7 to 14 basic. It is calculated as

pH = -log (H30+)

pH= -log (4x10-9M)

<em>pH=8,40</em>

8 0
3 years ago
What is the mass of 0.50 mole of copper?<br> 63.5 grams<br> O 32 grams<br> 3.2 grams<br> 2.18 grams
Amanda [17]

Answer:

32 g Cu

Explanation:

1 mol Cu    -> 63.5 g

0.5 mol Cu ->x

x=(0.5 mol *63.5 g)/1 mol      x= 32 g Cu

3 0
3 years ago
Why edible oils shows rancidity when stored for a long time
iren2701 [21]

When edible oils are idle and stored for a long amount of time, they undergo oxidation due to the exposure to oxygen. This oxidation causes rancidity in oils.

4 0
3 years ago
You need to prepare 1 L of the citric acid/citrate buffer. You have chosen to use Method 1 (see lab presentation). Calculate the
prisoha [69]

Answer:

3.11 is the pH of the buffer

Explanation:

The pH of a buffer is obtained using H-H equation:

pH = pKa + log [Conjugate base] / [Weak acid]

<em>Where pH is the pH of the buffer, pKa = -log Ka = 3.14 for the citric buffer and [] could be taken as the moles of each species.</em>

The citric acid,HX (Weak acid), reacts with NaOH to produce sodium citrate, NaX (weak base) and water:

HX + NaOH → H2O + NaX

That means the moles of NaOH added = Moles of sodium citrate produced

And the resulitng moles of HX = Initial moles - Moles NaOH added

<em>Moles HX and NaX:</em>

Moles NaOH = 0.100L * (0.65mol / L) = 0.065 moles NaOH = Moles NaX

Moles HX = 0.300L * (0.45mol / L) = 0.135 moles HX - 0.065 moles NaOH = 0.070 moles HX

Replacing in H-H equation:

pH = 3.14 + log [0.065mol] / [0.070mol]

pH = 3.11 is the pH of the buffer

8 0
3 years ago
How many molecules (not moles of nh3 are produced from 5.01×10?4 g of h2?
Irina-Kira [14]
2,02g     -------    6,02×10²³
5,01×10⁴g  ---    x

x=\frac{5,01*10^{4}g*6,02*10^{23}}{2,02g}=14,93*10^{27}

N₂         +        3H₂       ⇒           2NH₃
1mol      :        3mol        :           2mol
                       18,06×10²³  :       12,04×10²³
                       14,93×10²⁷  :        y

y=\frac{14,93*10^{27}*12,04*10^{23}}{18,06*10^{23}}\approx9,95*10^{27}
5 0
3 years ago
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