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7nadin3 [17]
3 years ago
8

Hydrazine, N2H4, is a corrosive liquid used in rocket and jet fuels. Ammonia, NH3, is a gas that dissolves in water to form a so

lution that can be used as a cleaning agent. How do hydrazine and ammonia illustrate the law of multiple proportions?
(the law of mulitple proportions is; statement that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.)
Chemistry
1 answer:
Lyrx [107]3 years ago
5 0

OK in the case of hydrazine  14 grams of nitrogen  combine with 2 gram of hydrogen and with ammonia 14 grams combine with 3 grams of hydrogen.

Ratio 2:3
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finlep [7]
The correct option in here is the first one: <span>pentane-1,5-diamine. This is also known as cadaverine. This is a common compund that produces that unpleasant odor from the fish. </span>
5 0
2 years ago
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jek_recluse [69]

Answer:

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Explanation:

had this ln a test and got it right

6 0
2 years ago
For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ
Komok [63]

Answer:

Check the explanation

Explanation:

Answer – Given, H_3PO_4 acid and there are three Ka values

K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}

The transformation of H_2PO_4- (aq) to HPO_4^2-(aq)is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

                             = 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

                             = 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

       = -log 6.2x10-8

       = 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

     = 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

6 0
2 years ago
what volume of co2 is produced at stp when 270g of glucose are consumed in the following reaction? c6h12o6 + 6o2(g) -&gt; 6co2 (
lana [24]

Answer:

202 L

Explanation:

Step 1: Write the balanced equation

C₆H₁₂O₆ + 6 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(l)

Step 2: Calculate the moles corresponding to 270 g of C₆H₁₂O₆

The molar mass of C₆H₁₂O₆ is 180.16 g/mol.

270 g × 1 mol/180.16 g = 1.50 mol

Step 3: Calculate the moles of CO₂ generated from 1.50 moles of glucose

The molar ratio of C₆H₁₂O₆ to CO₂ is 1:6. The moles of CO₂ formed are 6/1 × 1.50 mol = 9.00 mol

Step 4: Calculate the volume of 9.00 moles of CO₂ at STP

The volume of 1 mole of an ideal gas at STP is 22.4 L.

9.00 mol × 22.4 L/mol = 202 L

4 0
3 years ago
Which term describes this reaction?
DedPeter [7]

Answer:

C.

Explanation:

This is a reaction of elimination, because the water was removed and because of it double bond is formed.

7 0
2 years ago
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