All categories

2 years ago

What does he want? change into passive

Chemistry

1 answer:

Marina CMI [18]2 years ago

3 0

There’s no picture if you meant to add a picture

You might be interested in

CNC NaCl HCL are all classified as

Leona [35]

<span>A) elements and compounds. Compounds are two combined elements.</span>

6 0

2 years ago

Obtain a box of breakfast cereal and read the list of ingredients. What are four chemicals from the list? a. monoglycerides b. c

GREYUIT [131]

Answer:

B. cocamide DEA

C. folic acid

D. iron

G. lauryl glucoside

5 0

3 years ago

Leave the answer blank if no precipitate will form. (Express your answer as a chemical formula.) Formula of precipitate ZnCl2(aq

Rasek [7]

Answer:

1. Zn(OH)₂ (s)

2. Ag₂CO₃ (s)

3. Ni₃(PO₄)₂(s)

4. No reaction

5. (NH₄)₂CO₃(s)

Explanation:

Let's state the equations and we analyse some solubility and precipitation information:

ZnCl₂(aq) + 2KOH(aq) → Zn(OH)₂ (s) + 2KCl (aq)

All the salts from the halogens with group 1, are soluble.

The OH⁻ reacts to Zn cation in order to produce a precipitate. This is ok, but if the base is in excess, the Zn(OH)₂ will be soluble

K₂CO₃(aq) + 2AgNO₃(aq) → Ag₂CO₃ (s) ↓ + 2KNO₃(aq)

All salts from nitrate are soluble

All salts from carbonates are insoluble

2(NH₄)₃PO₄(aq) + 3Ni(NO₃)₂(aq) → Ni₃(PO₄)₂(s) ↓ + 6NH₄NO₃(aq)

Salts from phosphates are insoluble

All salts from nitrate are soluble

NaCl(aq) + KNO3(aq) → NO REACTION

All salts from nitrate are soluble

All the salts from the halogens with group 1, are soluble

Na₂CO₃(aq) + 2NH₄Cl(aq) → 2NaCl(aq) + (NH₄)₂CO₃(s) ↓

All salts from carbonates are insoluble

All the salts from the halogens with group 1, are soluble

7 0

2 years ago

Read 2 more answers

The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe

Irina-Kira [14]

Answer:

$t=147.24\ min$

Explanation:

Given that:

Half life = 30 min

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{30}\ min^{-1}$

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration $[A_0]$ = 7.50 mg

Final concentration $[A_t]$ = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

$0.25=7.50e^{-0.0231\times t}$

$750e^{-0.0231t}=25$

$x=\frac{\ln \left(30\right)}{0.0231}$

$t=147.24\ min$

6 0

3 years ago

What is the balanced chemical equation for the reaction of potassium with sulfur?

Aneli [31]

K + S = K2S

Potassium reacts with sulfur to produce potassium sulfide

8 0

3 years ago

What does he want? change into passive ​

What does he want? change into passive