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WINSTONCH [101]

3 years ago

6

Enter the molecular equation representing aqueous nitric acid and aqueous ammonia reacting. express your answer as a balanced mo

lecular equation. identify all of the phases in your answer.

2 answers:

NemiM [27]3 years ago

8 0

Answer : The balanced chemical reaction will be,

$HNO_3(aq)+NH_3(aq)\rightarrow NH_4NO_3(aq)$

Explanation :

Balanced chemical reaction : It is a reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

As per question.

When aqueous nitric acid react with aqueous ammonia then it gives ammonium nitrate as a product.

The balanced chemical reaction will be,

$HNO_3(aq)+NH_3(aq)\rightarrow NH_4NO_3(aq)$

This reaction is an acid-base reaction in which an acid react with a base to give salt and water as a product.

PtichkaEL [24]3 years ago

4 0

Aqueous nitric acid and aqueous ammonia reacts to form ammonium nitrate
HNO₃(aq)+NH₃(aq) = NH₄NO₃(aq)
HNO₃ +NH3 = NH₄ (+) + NO₃ (-)
Therefore the net ionic equation will be;
H⁺(aq) + NH₃ = NH₄⁺ (aq)

You might be interested in

What is the electron configuration for<br> 08<br> 16

Elis [28]

The electron configuration for Oxygen : [He] 2s²2p⁴

<h3>Further explanation</h3>

Writing electron configurations starts from the lowest to the highest sub-shell energy level. There are 4 sub-shells in the shell of an atom, namely s, p, d, and f. The maximum number of electrons for each sub-shell is

• s: 2 electrons

• p: 6 electrons

• d: 10 electrons and

• f: 14 electrons

Charging electrons in the sub-shell uses the following sequence:

<em>1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc. </em>

The element is Oxgen, with symbol O, and :

the atomic number=8=number of electron

the atomic mass=16

The electron configuration based on the number of electrons(for Oxygen=8), so the configuration :

$\tt _8^{16}O:1s^22s^22p^4$ or we can write with noble gas [He] 2s²2p⁴

3 0

2 years ago

If a system has a reaction quotient of 2.13 ✕ 10−15 at 100°C, what will happen to the concentrations of COBr2, CO, and Br2 as th

qaws [65]

This is an incomplete question, here is a complete question.

Consider the following equilibrium at 100°C.

$COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)$

$K_c=4.74\times 10^4$

Concentration at equilibrium:

$[COBr_2]=1.58\times 10^{-6}M$

$[Co]=2.78\times 10^{-3}M$

$[Br_2]=2.51\times 10^{-5}M$

If a system has a reaction quotient of 2.13 × 10⁻¹⁵ at 100°c, what will happen to the concentrations of COBr₂, Co and Br₂ as the reaction proceeds to equilibrium?

Answer : The concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

$COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)$

The expression for reaction quotient will be :

$Q=\frac{[CO][Br_2]}{[COBr_2]}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$Q=\frac{(2.78\times 10^{-3})\times (2.51\times 10^{-5})}{(1.58\times 10^{-6})}=4.42\times 10^{-2}$

The given equilibrium constant value is, $K_c=4.74\times 10^4$

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When $Q>K_c$ that means product > reactant. So, the reaction is reactant favored.

When $Q$ that means reactant > product. So, the reaction is product favored.

When $Q=K_c$ that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the $Q$ that means product < reactant. So, the reaction is product favored that means reaction must shift to the product (right) to be in equilibrium.

Hence, the concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

3 0

3 years ago

How many atoms of each element are contained in a single formula unit of Iron III formate, Fe(CH02)3 x H20? What percentrage by

dedylja [7]

The percentage by mass of the compound of water is going to be 8.62%.

7 0

3 years ago

What creates Earth’s magnetic field?

DIA [1.3K]

Earth's magnetic field doesn't have much to do with tectonic plates. But both are caused by movement of the magma. The magnetic field is caused by electrical currents flowing in the magma, roughly parallel with the Equator. Tectonic plates move around the surface of the Earth over the liquid magma.

So I'd go with B. mantle convection currents

5 0

3 years ago

An unknown compound with a molar mass of 223.94 g/mol consists of 32.18% c, 4.50% h, and 63.32% cl. find the molecular formula f

Dima020 [189]

The actual number of atoms of each element present in the molecule of the compound is represented by the formula known as molecular formula.

Molar mass of the unknown compound = 223.94 g/mol (given)

Mass of each element present in the unknown compound is determined as:

Mass of carbon, $C$ :

$\frac{32.18}{100}\times 223.94 = 72.06 g$

Mass of hydrogen, $H$ :

$\frac{4.5}{100}\times 223.94 = 10.08 g$

Mass of chlorine, $Cl$ :

$\frac{63.32}{100}\times 223.94 = 141.79 g$

Now, the number of each element in the unknown compound is determined by the formula:

$number of moles = \frac{given mass}{molar mass}$

Number of moles of $C$ :

$number of moles = \frac{72.06}{12} = 6.005 mole\simeq 6 mole$

Number of moles of $H$ :

$number of moles = \frac{10.08}{1} = 10.08 mole\simeq 10 mole$

Number of moles of $Cl$

$number of moles = \frac{141.79}{35.5} = 3.99 mole\simeq 4 mole$

Dividing each mole with the smallest number of mole, to determine the empirical formula:

$C_{\frac{6}{4}}H_{\frac{10}{4}}Cl_{\frac{10}{4}}$

$C_{1.5}H_{2.5}Cl_{1}$

Multiplying with 2 to convert the numbers in formula into a whole number:

So, the empirical formula is $C_{3}H_{5}Cl_{2}$ .

Empirical mass = $12\times 3+1\times 5+2\times 35.5 = 112 g/mol$

In order to determine the molecular formula:

n = $\frac{molar mass}{empirical mass}$

n = $\frac{223.94}{112} = 1.99 \simeq 2$

So, the molecular formula is:

$2\times C_{3}H_{5}Cl_{2} = C_{6}H_{10}Cl_{4}$

6 0

3 years ago