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Leto [7]
3 years ago
15

Pleaseeeee hellppppp !!!!

Chemistry
1 answer:
Tomtit [17]3 years ago
3 0

Answer:

  1. both
  2. cation
  3. anion
  4. both
  5. cation
  6. anion
  7. both
  8. both
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What are the products of burning fuel
aleksley [76]
<span>water vapour.
carbon dioxide.
carbon monoxide.particles.
<span>sulfur dioxide.</span></span>
5 0
3 years ago
Which statement is true? Postively charged objects attract other positively charged objects negatively charged objects attract o
ANEK [815]

Answer:

Explanation:one is true because if it is positiveit would not be attracting it would be sepperating

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3 years ago
Select the reagents you would use to synthesize the compounds below from benzene. Use the minimum number of steps. No more than
jekas [21]

Answer:

Explanation:

Using the necessary reagents to faciliate the synthesis of the organic compounds as shown in the attached file.

8 0
3 years ago
A 5.5 g sample of a substance contains only carbon and oxygen. Carbon makes up 35% of the mass of the substance. The rest is mad
Umnica [9.8K]

We have been given the condition that carbon makes up 35% of the mass of the substance and the rest is made up of oxygen. With this, it can be concluded that 65% of the substance is made up of oxygen. If we let x be the mass of oxygen in the substance, the operation that would best represent the scenario is,

<span>                                       x = (0.65)(5.5 g)</span>

<span>                                       <em> </em><span><em>x = 3.575 g</em></span></span>

8 0
3 years ago
The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentrati
a_sh-v [17]

Answer : The concentration of NOBr after 95 s is, 0.013 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.80M^{-1}s^{-1}

t = time taken  = 95 s

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.86 M

Now put all the given values in above equation, we get:

0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)

[A] = 0.013 M

Hence, the concentration of NOBr after 95 s is, 0.013 M

4 0
3 years ago
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