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labwork [276]
4 years ago
11

Complete and balance the equation for this single-displacement reaction. Phases are optional. Li+NaOH-->

Chemistry
2 answers:
sdas [7]4 years ago
6 0
The complete and <span>balanced equation for this single-displacement reaction would be written as follows:

</span><span>Li+NaOH--> LiOH + Na

In this reaction, lithium replace sodium in the compound sodium hydroxide forming lithium hydroxide and sodium metal as products. Hope this answers the question.</span>
andre [41]4 years ago
4 0

Answer: Li+NaOH\rightarrow LiOH+Na

Explanation:

A single replacement reaction is one in which a more reactive element displaces a less reactive element from its salt solution.

The balanced chemical equation is :

Li+NaOH\rightarrow LiOH+Na

This is a single replacement reaction in which lithium replaces sodium from its aqueous solution. As lithium is more reactive than sodium, it easily loses electrons to convert to Li^{+} in LiOH and Na^+ in NaOH accepts electrons to form Na.

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Match each description below:
neonofarm [45]

Answer:

a) will react with water: K

b) will react with steam or acid, but not water: Cr & Sn

d) will not react with water, steam, or acid: Cu

Explanation:

K reacts violently with water.

Cr will react with steam to form an oxide + H gas and will react with most acids

Sn will react with steam to form SnO₂ + H gas and, though it does not react as rapidly as other metals in acid, it dissolves easily in concentrated acids

Cu is an extremely nonreactive metal, which is what makes it so suitable for wiring

5 0
3 years ago
A galvanic (voltaic) cell has the generic metals X and Y as electrodes. X is more reactive than Y , that is, X more readily reac
koban [17]

Answer:

(a) X electrode

(b) Y electrode

(c) Y electrode

(d) X electrode

(e) Y electrode

Explanation:

<em>A galvanic (voltaic) cell has the generic metals X and Y as electrodes. X is more reactive than Y, that is, X more readily reacts to form a cation than Y does.</em>

In the X electrode occurs the oxidation whereas in the Y electrode occurs the reduction.

Oxidation: X(s) → X⁺ⁿ(aq) + n e⁻

Reduction: Y⁺ˣ(aq) + x e⁻ → Y(s)

<em>Classify the descriptions by whether they apply to the X or Y electrode. </em>

<em>(a) anode.</em> Is where the oxidation takes place (X electrode).

<em>(b) cathode.</em> Is where the reduction takes place (Y electrode).

<em>(c) electrons in the wire flow toward.</em> Electrons in the wire flow toward the cathode (Y electrode).

<em>(d) electrons in the wire flow away.</em> Electrons in the wire flow away from the anode (X electrode).

<em>(e) cations from salt bridge flow toward.</em> Cations from the salt bridge flow toward the cathode (Y electrode) to maintain the electroneutrality.

5 0
3 years ago
Pure chlorobenzene (C6H5Cl) has a normal boiling point of 131.00 °C. A solution of 32.5 g of 2,8-dibromodibenzofuran (C12H6Br2O)
vichka [17]

Answer:

Kb →  1.56 °C / m

Explanation:

This is all about boiling point elevation, the colligative property that shows that boiling point for a solution is higher than boiling point of pure solvent.

This is the formula: ΔT = Kb . m . i

where i is the Van't Hoff factor (ions dissolved in solution). As these are organic compounds, we assume they are non electrolytic,

m is molality (mol of solute / 1kg of solvent)

Kb is our unknown. The value for ebulloscopic constant, it is specific for each solvent.

ΔT = T° boiling from solution - T° boiling from solute

First of all, let's determine the moles of solute.

Mass / Molar mass → 32.5 g/ 113.45 g/mol = 0.286 mol

Molality is mol of solute/ 1 kg of solvent

We must convert the mass from g to kg

195g . 1kg /1000 = 0.195 kg

Molality = 0.286 mol / 0.195 kg = 1.47 m

Let's replace the values in the formula

133.30 °C - 131°C = Kb . 1.47m .1

2.30°C / 1.47 m =  Kb →  1.56 °C / m

3 0
3 years ago
Use the following equation to complete the questions below. 2C2H6 + 7O2 → 4CO2 + 6H2O How many moles of oxygen would you need to
worty [1.4K]
43.75 moles of oxygen and 27.42 moles of water.

4 0
3 years ago
The mass of an atom of beryllium is 1.5 x 10^-26 kg. How many beryllium atoms are present in beryllium film of mass 0.37 g used
timama [110]

Answer:

There are 2.5 * 10^22 beryllium atoms in 0.37 grams

Explanation:

Step 1: Data given

The mass of an atom of beryllium is 1.5 x 10^-26 kg.

Total mass = 0.37 grams

Step 2: Calculate number of atoms

1 atom = 1.5 * 10 ^-26 kg

X atoms = 0.00037 kg

X = 0.00037/1.5*10^-26

X = 2.467 * 10^22 atoms ≈ 2.5 * 10^22 atoms

There are 2.5 * 10^22 beryllium atoms in 0.37 grams

8 0
3 years ago
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