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sertanlavr [38]
3 years ago
14

A 100. ml portion of 0.250 m calcium nitrate solution is mixed with 400. ml of 0.100 m nitric acid solution. what is the final c

oncentration of the nitrate ion?
Chemistry
1 answer:
pshichka [43]3 years ago
6 0
<span>Answer: Moles Ca(NO3)2 = 100 x 0.250 / 1000 = 0.025 Ca(NO3)2 >> Ca2+ + 2NO3- Moles NO3- = 2 x 0.025 = 0.05 Moles HNO3 = 400 x 0.100 / 1000 = 0.04 Total moles = 0.05 + 0.04 = 0.09 Total volume = 500 ml = 0.500 L M = 0.09 / 0.500 = 0.18</span>
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An unknown gaseous substance has a density of 1.06 g/L at 31 °C and 371 torr. If the substance has the following percent composi
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Answer:

C) C4H6 - Right answer

Explanation:

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P . V = n . R . T

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Mass / Molar mass = Moles → That's why the 1.06 g / MM

0.0195 mol = 1.06g / MM

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Now, let's use the composition

100 g of compound have 88.8 g of C

54.3 g of compound have ___ (54.3  . 88.8) /100 = 48 g of C

100 g of compound have 11.2 g of H

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48 g of C are included un 4 atoms

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4 0
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Explanation:

4 0
2 years ago
Someone help me please!!!
Montano1993 [528]

A. 6 moles

B. 9 moles

C. 3 moles

D.  20 moles

I think please check me, in case I am wrong

8 0
3 years ago
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