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ss7ja [257]
3 years ago
11

What monatomic ions would you expect potassium (Z = 19) and bromine (Z = 35) to form?

Chemistry
1 answer:
timofeeve [1]3 years ago
6 0

Answer:

Explanation:

Potassium ion will be formed when potassium atom loses one electron to become positively charged

K₁₉ →→  K₁₉⁺ + e⁻

Bromine ion is formed by gain of electron

Br₃₅ + e⁻ → Br₃₅⁻

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How do you find the metal “M”
Elodia [21]

M = Sr (strontium)

<em>Step 1.</em> Calculate the <em>moles of CO_2</em>.

Moles of CO_2 = 0.395 g CO_2 × (1 mol CO_2/44.01 g CO_2)

= 0.008 975 mol CO_2

<em>Step 2</em>. Calculate the <em>moles of MCO_3</em>.

Moles of MCO_3 = 0.008 975 mol CO_2 × (1 mol MCO_3/1 mol CO_2)

= 0.008 975 mol MCO_3

<em>Step 3</em>. Calculate the molar mass of <em>MCO_3</em>

MM = grams/moles = 1.324 g/0.008 75 mol = 147.5 g/mol

<em>Step 4</em>. Calculate the <em>atomic mass of M</em>

M_r = <em>x</em> + 12.01 + 3×16.00 = <em>x</em> + 60.01 = 147.5

<em>x</em> = 147.5 – 60.01 = 87.5

<em>Step 5. Identify M</em>.

The element with the closest atomic mass is Sr (A_r = 87.6).

∴ M = Sr and the compound is SrCO_3.

5 0
3 years ago
A piston confines 0.200 mol Ne(g) in 1.20 at 25 degree C. Two experiments are performed. (a) The gas is allowed to expand throug
djyliett [7]

Answer:

The second experiment (reversible path) does more work

Explanation:

Step 1:

A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C

<em>(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm</em>

<em></em>

Irreversible path: w =-Pex*ΔV

⇒ with Pex = 1.00 atm

⇒ with ΔV = 1.20 L

W = -(1.00 atm) * 1.20 L

W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J

<em>(b) The gas is allowed to expand reversibly and isothermally to the same final volume.</em>

<em></em>

W = -nRTln(Vfinal/Vinitial)

⇒ with n = the number of moles = 0.200

⇒ with R = gas constant = 8.3145 J/K*mol

⇒ with T = 298 Kelvin

⇒ with Vfinal/Vinitial  = 2.40/1.20 = 2

W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)

W = -343.5 J

The second experiment (reversible path) does more work

7 0
3 years ago
What mass of octane must be burned in order to liberate 5270 kj of heat? δhcomb = -5471 kj/mol?
katrin2010 [14]
As,
                           5471 kJ heat is given by  =  1 mole of Octane
Then,
                    5310 kJ heat will be given by  = X moles of Octane

Solving for X,
                                  X  =  (5310 kJ × 1 mol) ÷ 5471 kJ

                                  X  =  0.970 moles of Ocatne

So, 0.970 moles of Octane will liberate 5310 kJ energy. Now changing moles to mass,
As,
                                  Moles  =  mass / M.mass
Or,
                                  Mass  =  Moles × M.mass
Putting values,
                                  Mass  =  0.970 mol × 114.23 g/mol

                                  Mass  =  110.83 g of Octane
6 0
3 years ago
Joel has noticed that his engine is not compressing properly during the compression and ignition step. What problem will this po
denis-greek [22]

The answer is: The engine will run inefficiently APEX....

7 0
3 years ago
Identify which statement describes the following process. 2Ce + 3Cu2+ → 3Cu + 2Ce3+
Georgia [21]
Answer: <span>A-Ce is oxidized because it is losing electrons and Cu is reduced because it is gaining electrons</span><span>.

</span>There are two reactions in the equation, oxidation and reduction. A molecule that oxidized will lose electrons while the molecule that reduced will gain electrons. In this case, Cu2+ changed into Cu which means its oxidation number reduced from +2 into 0. Ce oxidation number increased from 0 into +3 
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3 years ago
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