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3 years ago

Which reaction is the opposite of a decomposition reaction?

Chemistry

1 answer:

tresset_1 [31]3 years ago

5 0

Answer:

B) Synthesis reaction

Explanation:

synthesis reactions because they take apart larger molecules or compounds

Hope that helps

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SiCl4 + H2O = H4SiO4 + HCl

vichka [17]

Hey there!

Balance the equation:

SiCl₄ + H₂O → H₄SiO₄ + HCl

Balance H.

2 on the left, 5 on the right. Add a coefficient of 3 in front of H₂O and a coefficient of 2 in front of HCl.

SiCl₄ + 3H₂O → H₄SiO₄ + 2HCl

Balance O.

3 on the left, 4 on the right. Change the coefficient of 3 in front of H₂O to a 4.

SiCl₄ + 4H₂O → H₄SiO₄ + 2HCl

This unbalanced our H, so change the coefficient of 2 in front of HCl to a 4.

SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl

Balance Cl.

4 on the left, 4 on the right. Already balanced.

Balance Si.

1 on the left, 1 on the right. Already balanced.

Our final balanced equation:

SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl

Hope this helps!

8 0

3 years ago

If 97.3 L of NO2 forms measured at 35 C and 632 mm Hg. What is the percent yield?

enyata [817]

<span>a. Use PV = nRT and solve for n = number of mols O2.
mols NO = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR.

b.
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L.

</span>and % will be 60%.

3 0

3 years ago

Ethylenediamine (en) forms an octahedral complex with Ni2+(aq) with the formula [Ni(en)3]2+. Ni2+(aq) + 3 en ⇌ [Ni(en)3]2+(aq) K

yawa3891 [41]

Answer:

$3.125\times 10^{-19} mol/L$ is the concentration of $Ni^{2+}(aq)$ in the solution.

Explanation:

$Ni^{2+}(aq) + 3 en\rightleftharpoons [Ni(en)_3]^{2+}(aq)$

Concentration of nickel ion = $[Ni^{2+}]=x$

Concentration of nickel complex= $[[Ni(en)_3]^{2+}]=\frac{0.16 mol}{2 L}=0.08 mol/L$

Concentration of ethylenediamine = $[en]=\frac{0.80 mol}{2 L}=0.40 mol/L$

The formation constant of the complex = $K_f=4.0\times 10^{18}$

The expression of formation constant is given as:

$K_f=\frac{[[Ni(en)_3]^{2+}]}{[Ni^{2+}][en]^3}$

$4.0\times 10^{18}=\frac{0.08 mol/L}{x\times (0.40 mol/L)^3}$

$x=\frac{0.08 mol/L}{4.0\times 10^{18}\times (0.40 mol/L)^3}$

$x=3.125\times 10^{-19} mol/L$

$3.125\times 10^{-19} mol/L$ is the concentration of $Ni^{2+}(aq)$ in the solution.

6 0

3 years ago

Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly

trapecia [35]

Answer : The mass of silver sulfadiazine produced can be, 71.35 grams.

Solution : Given,

Mass of $Ag_2O$ = 25.0 g

Mass of $C_{10}H_{10}N_4SO_2$ = 50.0 g

Molar mass of $Ag_2O$ = 231.7 g/mole

Molar mass of $C_{10}H_{10}N_4SO_2$ = 250.3 g/mole

Molar mass of $AgC_{10}H_{9}N_4SO_2$ = 357.1 g/mole

First we have to calculate the moles of $Ag_2O$ and $C_{10}H_{10}N_4SO_2$ .

$\text{ Moles of }Ag_2O=\frac{\text{ Mass of }Ag_2O}{\text{ Molar mass of }Ag_2O}=\frac{25.0g}{231.7g/mole}=0.1079moles$

$\text{ Moles of }C_{10}H_{10}N_4SO_2=\frac{\text{ Mass of }C_{10}H_{10}N_4SO_2}{\text{ Molar mass of }C_{10}H_{10}N_4SO_2}=\frac{50.0g}{250.3g/mole}=0.1998moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Ag_2O(s)+2C_{10}H_{10}N_4SO_2(s)\rightarrow 2AgC_{10}H_9N_4SO_2(s)+H_2O(l)$

From the balanced reaction we conclude that

As, 2 mole of $C_{10}H_{10}N_4SO_2$ react with 1 mole of $Ag_2O$

So, 0.1998 moles of $C_{10}H_{10}N_4SO_2$ react with $\frac{0.1998}{2}=0.0999$ moles of $Ag_2O$

From this we conclude that, $Ag_2O$ is an excess reagent because the given moles are greater than the required moles and $C_{10}H_{10}N_4SO_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgC_{10}H_9N_4SO_2$

From the reaction, we conclude that

As, 2 mole of $C_{10}H_{10}N_4SO_2$ react with 2 mole of $AgC_{10}H_9N_4SO_2$

So, 0.1998 mole of $C_{10}H_{10}N_4SO_2$ react with 0.1998 mole of $AgC_{10}H_9N_4SO_2$

Now we have to calculate the mass of $AgC_{10}H_9N_4SO_2$

$\text{ Mass of }AgC_{10}H_9N_4SO_2=\text{ Moles of }AgC_{10}H_9N_4SO_2\times \text{ Molar mass of }AgC_{10}H_9N_4SO_2$

$\text{ Mass of }AgC_{10}H_9N_4SO_2=(0.1998moles)\times (357.1g/mole)=71.35g$

Therefore, the mass of silver sulfadiazine produced can be, 71.35 grams.

8 0

3 years ago

Delocalization which occurs through overlap between stigma and p orbitals

elixir [45]

Answer:

Single covalent bonds that form between nuclei are created from the "head-to-head" overlap of orbitals and are called stigma (s) bonds. ... Another type of bond, a pi (p) bond is formed when two p orbitals overlap. Pi bonds are found in double and triple bond structures

Explanation:

Delocalization happens when electric charge is spread over more than one atom. For example, bonding electrons may be distributed among several atoms that are bonded together.

7 0

3 years ago