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Mkey [24]
3 years ago
10

For each reaction below, use the drop-down menus to select which compound will form a precipitate. Note: It is possible that no

precipitates form.
Na2S + Cd(NO3)2 → CdS + 2NaNO3

FeCl3 + 3KOH → 3KCl + Fe(OH)3

Na2CO3 + Ba(NO3)2 → BaCO3 + 2NaNO3
Chemistry
1 answer:
Dimas [21]3 years ago
8 0

Answer:

1. Cadmium sulfide

2. Iron (III) hrydroxide

3. Barium carbonate

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The answer would be B.  Increased fertility downstream. Hope this helps. It's been a while since iv'e done the test, but I think this is the correct answer.

4 0
3 years ago
What is produced by a neutralization reaction between an arrhenius acid and an arrhenius base?
stepladder [879]

Answer: Salt and Water

Explanation:

An Arrhenius acid (HCl) can best be defined as any substance that when added to water increases the concentration of H+ ions.

While an Arrhenius base (KOH) is any substance that when added to water increases the concentration of OH- ions.

When an Arrhenius acid such as HCl reacts with an Arrhenius base such as KOH, the end products will be salt and water, in a process called Neutralization Reaction.

HCl (aq) + KOH (aq)  -------> KCl (aq) + H2O (l)

4 0
3 years ago
PLS HELP ASAP THANKS ILL GIVE BRAINLKEST PLS THANKS PLS ASAP PLS PLS HELP ASAP THANKS
Alborosie

Answer:

b

Explanation:

3 0
3 years ago
Select the single best answer. Explain why (CH3)2CHCH(Br)CH2CH3 reacts faster than (CH3)2CHCH2CH(Br)CH3 in an E2 reaction, even
goldenfox [79]

Answer:

The major and minor products formed from the first structure have more alkyl groups on the C═C than those formed from the second structure

Explanation:

When we consider the structures of the major and minor products from the dehydrohalogenation of structure 1 by E2 mechanism, it is easy to see that the both products are highly substituted by alkyl groups. It is a fact in organic chemistry that the more substituted an alkene is the more stable it is and the more quickly it is formed; Hence the answer

8 0
4 years ago
This shielding is a maximum for UV light having a wavelength of 295nanometers. What is the frequency in hertz of this particular
LUCKY_DIMON [66]

Answer:

\large \boxed{\text{1.016 $\times 10^{15}$ Hz}}

Explanation:

The formula relating the frequency (f) and wavelength (λ) is

fλ = c  

1. Convert nanometres to metres

295 nm = 295 × 10⁻⁹ m

2. Calculate the frequency

\begin{array}{rcl}f \times 295 \times 10^{-9} \text{ m}& =& 2.998 \times 10^{8} \text{ m$\cdot$s$^{-1}$}\\\\f & = & \dfrac{2.998 \times 10^{8} \text{ m$\cdot$s$^{-1}$}}{295 \times 10^{-9}\text{ m}}\\\\& = &1.016 \times 10^{15}\text{ s}^{-1}\\& = &\large \boxed{\textbf{1.016 $\mathbf{\times 10^{15}}$ Hz}}\end{array}

8 0
4 years ago
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