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2 years ago

Write a balanced chemical equation for the standard formation reaction of solid sodium hydrogen carbonate nahco3 .

1 answer:

5 0

There are a number of methods of which we can form sodium bicarbonate. This compound is commonly known as baking soda. It can be prepared from the reaction of sodium hydroxide and carbonic acid. Carbonic acid in water dissociates into hydrogen ions and the bicarbonate ion while sodium hydroxide would ionize into sodium ions and hydroxide ions. With this, these ions would react and form the sodium bicarbonate salt and water. The chemical reaction would be expressed as follows:

NaOH + H2CO3 = H2O + NaHCO3

Sodium bicarbonate is used in cooking, as a toothpaste and as a cleaning substance. Also, it is used in medical applications like for the preparation of the dialysate solution.

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A sample of element X contains 90% X-35 atoms, 8.0% X-37 atoms, and 2.0% X-38 atoms. The average atomic mass will be closest to

Ne4ueva [31]

To find average atomic mass you multiply the mass of each isotope by its percentage, and then add the values up.

35 * 0.90 + 37 * 0.08 + 38 * 0.02 = 35.22

Average atomic mass closest to 35.22 amu.

6 0

2 years ago

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A scuba tank contains a mixture of oxygen (O2) and nitrogen (N2) gas. The oxygen has a partial pressure of PO2=5.62MPa. The tota

dmitriy555 [2]

Answer:

21.16 MPa

Explanation:

Partial pressure of oxygen = 5.62 MPa

Total gas pressure = 26.78 MPa

But

Total pressure of the gas= sum of partial pressures of all the constituent gases in the system.

This implies that;

Total pressure of the system = partial pressure of nitrogen + partial pressure of oxygen

Hence partial pressure of nitrogen=

Total pressure of the system - partial pressure of oxygen

Therefore;

Partial pressure of nitrogen= 26.78 - 5.62

Partial pressure of nitrogen = 21.16 MPa

7 0

2 years ago

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The dipole moment (μ) of HBr (a polar covalent molecule) is 0.838D (debye), and its percent ionic character is 12.4 % . Estimate

myrzilka [38]

When two electrical charges, of opposite sign and equal magnitude, are separated by a distance, a dipole is established. The size of a dipole is measured by its dipole moment (μμ). Dipole moment is measured in Debye units, which is equal to the distance between the charges multiplied by the charge (1 Debye equals 3.34×10−30Cm3.34×10−30Cm). The dipole moment of a molecule can be calculated by Equation 1.11.1:

μ = qr

where

μ⃗ μ→ is the dipole moment vector qiqi is the magnitude of the ithith charge, and r⃗ ir→i is the vector representing the position of ithith charge.

r = μ/q

r = [0.838D(3.34×10−30 C⋅m/ 1D)]/ (1.6×10−19 C) *0.124
 r = 1.41 x10^-10 m

7 0

2 years ago

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What is the net ionic equation of the reaction of MgCl2 with NaOH? Express your answer as a chemical equation. View Available Hi

kari74 [83]

Answer:

Net ionic equation for the reaction between MgCl₂ and NaOH in water:

$\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s)$ .

Net ionic equation for the reaction between MgSO₄ and BaCl₂ in water:

$\rm {Ba}^{2+}\; (aq) + {SO_4}^{2-}\;(aq) \to BaSO_4\; (s)$ .

Explanation:

Start by finding the chemical equations for each reaction:

MgCl₂ reacts with NaOH to form Mg(OH)₂ and NaCl. This reaction is a double decomposition reaction (a.k.a. double replacement reaction, salt metathesis reaction.) This reaction is feasible because one of the products, Mg(OH)₂, is weakly soluble in water and exists as a solid precipitate.

$\rm MgCl_2\; (aq) + 2\; NaOH\; (aq)\to Mg(OH)_2 \; (s) + 2\; NaCl\; (aq)$ .

MgSO₄ reacts with BaCl₂ in a double decomposition reaction to produce BaSO₄ and MgCl₂. Similarly, the solid product BaSO₄ makes this reaction is feasible.

$\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq)$ .

How to rewrite a chemical equation to produce a net ionic equation?

Rewrite all reactants and products that ionizes completely in the solution as ions.
Eliminate ions that exist on both sides of the equation to produce a net ionic equation.

Typical classes of chemicals that ionize completely in water:

Soluble salts,
Strong acids, and
Strong bases.

Keep the formula of salts that are not soluble in water, weak acids, weak bases, and water unchanged.

Take the first reaction as an example, note the coefficients:

MgCl₂ is a salt and is soluble in water. Each unit of MgCl₂ can be written as $\rm Mg^{2+}$ and $\rm 2\; Cl^{-}$ .
NaOH is a strong base. Each unit of NaOH can be written as $\rm Na^{+}$ and $\rm OH^{-}$ .
Mg(OH)₂ is a weak base and should not be written.
NaCl is a salt and is soluble in water. Each unit of NaCl can be written as $\rm Na^{+}$ and $\rm Cl^{-}$ .

$\rm Mg^{2+} + 2\; Cl^{-} + 2\; Na^{+} + 2\; OH^{-} \to Mg(OH)_2\;(s) + 2\; Na^{+} + 2\; Cl^{-}$ .

Ions on both sides of the equation:

$\rm 2\; Cl^{-}$ , and
$\rm 2\; Na^{+}$ .

Add the state symbols:

$\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s)$ .

For the second reaction:

$\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq)$ .

$\rm Mg^{2+} + 2\; {SO_4}^{2-} + Ba^{2+} + 2\; Cl^{-} \to BaSO_4\; (s) + Mg^{2+} + 2\; Cl^{-}$ .

$\rm Ba^{2+}\; (aq) + {SO_4}^{2-}\; (aq) \to BaSO_4\; (s)$ .

5 0

2 years ago

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Part a how many grams of xef6 are required to react with 0.579 l of hydrogen gas at 6.46 atm and 45°c in the reaction shown belo

Nina [5.8K]

First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.

PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂

The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = 11.69 g

6 0

2 years ago