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prisoha [69]
3 years ago
7

4. Which is produced when a base reacts with water?

Chemistry
2 answers:
a_sh-v [17]3 years ago
6 0

Answer:

OH^{-1}  ions.

Explanation:

Hydrogen ions (H+) are attracted to the negative oxygen end of a water molecule, combining to form hydronium ions.

Base: Very definition of base is that it produces OH- on dissolution with water. Bases have a pH greater than 7. A base is any substance that dissolves in water and produces hydroxide ions (OH-).

Dennis_Churaev [7]3 years ago
5 0

Answer:

hydroxide ion

Explanation:

I took the quiz

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Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and
rusak2 [61]

Answer:

4Zn_(_s_)~+~7OH^-~_(_a_q_)~+~NO_3^-_(_a_q_)~+~6H_2O_(_l_)~-->4Zn(OH)_4^-^2_(_a_q_)~+~NH_3_(_g_)

-) Oxidizing agent: NO_3^-_(_a_q_)

-) Reducing agent: Zn_(_s_)

Explanation:

The first step is separate the reaction into the <u>semireactions</u>:

A.Zn~->Zn(OH)_4^-^2

B.NO_3^-~->~NH_3

If we want to balance in <u>basic medium </u>we have to follow the rules:

1. We adjust the oxygen with OH^-

2. We adjust the H with H_2O

3. We adjust the charge with e^-

Lets balance the first semireaction A. :

Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-

Now, lets balance semireaction B:

NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-

Finally, we have to add the two semireactions:

_________________________________________

8~(Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-)

2~(NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-)

_________________________________________

(8Zn~+~32OH^-~->8Zn(OH)_4^-^2~+~16e^-)

(2NO_3^-~+~16e^-~+~12H_2O~->~2NH_3~+~18OH^-)

Cancel out the species on both sides:

8Zn~+~14OH^-~+~2NO_3^-~+~12H_2O~-->8Zn(OH)_4^-^2~+~2NH_3

Simplifying the equation :

4Zn~+~7OH^-~+~NO_3^-~+~6H_2O~-->4Zn(OH)_4^-^2~+~NH_3

The Zn_(_s_) is <u>oxidized</u> therefefore is the <u>reducing agent</u>. The NO_3^-_(_a_q_)is<u> reduced</u> therefore is the <u>oxidizing agent</u>.

4 0
4 years ago
Ribosomes are cell structures that make
Vitek1552 [10]

Answer:

Ribosomes preform biological synthesis. (mRNA Translation). They also link amino acids together to form polypeptide chain.

8 0
3 years ago
Read 2 more answers
60 N net force is applied to a 12 KG box. What is the acceleration of the box?
galben [10]

Answer:

The answer is

<h2>5 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and it's net force can be found by using the formula

acceleration =  \frac{force}{mass}  \\

From the question

force = 60 N

mass = 12 kg

So we have

acceleration =  \frac{60}{12}  \\

We have the final answer as

<h3>5 m/s²</h3>

Hope this helps you

8 0
4 years ago
An -ate or -ite at the end of a compound name usually indicates that the compound contains *
nikdorinn [45]

Explanation:

<h2>1.-Ate-contains 3 elements in which one is oxygen</h2><h2>2.-ite-contains 3 elements in which one is Oxygen but the oxygen is less than that of -Ate</h2><h2>3.-ide has 2 elements</h2>
3 0
3 years ago
(6) Compare a CSTR with a PFR below. a. A flow of 0.3 m3/s enters a CSTR (volume of 200 m3) with an initial concentration of spe
Dmitry [639]

Answer:

Explanation:

Given that:

The flow rate Q = 0.3 m³/s

Volume (V) = 200 m³

Initial concentration C_o = 2.00 ms/l

reaction rate K = 5.09 hr⁻¹

Recall that:

time (t) = \dfrac{V}{Q}

time (t) = \dfrac{200}{0.3}

time (t) = 666.66 \ sec

time (t) = \dfrac{666.66 }{3600} hrs

time (t) = 0.185 hrs

\text{Using First Order Reaction:}

\dfrac{dc}{dt}=kc

where;

t = \dfrac{1}{k} \Big( \dfrac{C_o}{C_e}-1 \Big)

0.185 = \dfrac{1}{5.09} \Big ( \dfrac{200}{C_e}- 1 \Big)

0.942 =  \Big ( \dfrac{200}{C_e}- 1 \Big)

1+ 0.942 =  \Big ( \dfrac{200}{C_e} \Big)

\dfrac{200}{C_e} = 1.942

C_e = \dfrac{200}{1.942}

\mathbf{C_e = 102.98 \ mg/l}

Thus; the concentration of species in the reactant = 102.98 mg/l

b). If the plug flow reactor has the same efficiency as CSTR, Then:

t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} =0.196(0.663)

\dfrac{V_{PFR}}{0.3 hrs} =0.196(0.663)

\dfrac{V_{PFR}}{0.3*3600 sec} =0.196(0.663)

V_{PFR} =0.196(0.663)*0.3*3600

V_{PFR} = 140.34 \ m^3

The volume of the PFR is ≅ 140 m³

3 0
3 years ago
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