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solmaris [256]
2 years ago
11

Select the compound that is soluble in water?

Chemistry
1 answer:
Aliun [14]2 years ago
6 0

Answer:

NaOH

Explanation:

NaOH is among the water soluble bases

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0.415 g of an unknown triprotic acid are used to make a 100.00 mL solution. Then 25.00 mL of this solution is transferred to an
Arturiano [62]

Answer:

Explanation:

Initial burette reading = 1.81 mL

final burette reading = 39.7 mL

volume of NaOH used = 39.7 - 1.81 = 37.89 mL .  

37.89 mL of .1029 M NaOH is used to neutralise triprotic acid

No of moles contained by 37.89 mL of .1029 M NaOH

= .03789 x .1029 moles

= 3.89 x 10⁻³ moles

Since acid is triprotic ,  its equivalent weight = molecular weight / 3

No of moles of triprotic acid = 3.89 x 10⁻³ / 3

= 1.30   x 10⁻³ moles .

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3 years ago
Speed is the ratio between distance and <br><br> Pls answer ASP!!
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Answer: Explanation: Speed is simply velocity without a specified direction (speed is the magnitude of the velocity vector). Regardless of the two measurements, they both relate distance to time, but are slightly different. Instantaneous speed is the derivative of the total distance covered with respect to time.

Explanation: I hope this helps!

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Calculate the frequency of a yellow light with a wavelength of 5.60 x 10-9 m.
Whitepunk [10]

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47

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By sprouting in the rainforest canopy, the strangler fig seedling is exposed to more
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8 0
2 years ago
3.) 2.46 grams of Cu(NO3)2 would contain how many formula units?
RUDIKE [14]

Answer:

7.90×10²¹ formula units

Explanation:

From the question given above, the following data were obtained:

Mass of Cu(NO₃)₂ = 2.46 g

Formula units of Cu(NO₃)₂ =?

From Avogadro's hypothesis,

1 mole of Cu(NO₃)₂ = 6.02×10²³ formula units

Next, we shall determine the mass of 1 mole of Cu(NO₃)₂. This can be obtained as follow:

1 mole of Cu(NO₃)₂ = 63.5 + 2[14 + (3×16)]

= 63.5 + 2[14 + 48]

= 63.5 + 2[62]

= 63.5 + 124

= 187.5 g

Thus,

187.5 g of Cu(NO₃)₂ = 6.02×10²³ formula units

Finally, we shall determine the formula units contained in 2.46 g of Cu(NO₃)₂. This can be obtained as follow:

187.5 g of Cu(NO₃)₂ = 6.02×10²³ formula units.

Therefore,

2.46 g of Cu(NO₃)₂ =

(2.46 × 6.02×10²³)/187.5

= 7.90×10²¹ formula units

Thus, 2.46 g of Cu(NO₃)₂ contains 7.90×10²¹ formula units

7 0
2 years ago
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