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densk [106]
3 years ago
7

Which group of elements will readily lose one electron to meet the octet rule? Alkaline Earth Metals c. Halogens Alkali Metals d

. Both A and B
Chemistry
1 answer:
vova2212 [387]3 years ago
4 0

Answer:

Alkali metals

Explanation:

The alkali metals are a group of metallic elements which are present in the first group of the periodic table. In other words, they are present in group 1 of the periodic table. These elements have one electron in their valence shell, the reason why they are placed in group 1.

They ionize by losing one electron to achieve the configuration of the nearest Noble gas or inert gas. Because they need to offset only one electron in their outermost shell, they are very chemically reactive and hence rarely occur in the free state.

Examples of elements in this group include lithium, potassium and sodium. They each have one electron only in their outermost shells.

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Suppose 550.mmol of electrons must be transported from one side of an electrochemical cell to another in 49.0 minutes. Calculate
nekit [7.7K]

Answer:

18.0 Ampere is the size of electric current that must flow.

Explanation:

Moles of electron , n = 550 mmol = 0.550 mol

1 mmol = 0.001 mol

Number of electrons = N

N=N_A\times n

Charge on N electrons : Q

Q = N\times 1.602\times 10^{-19} C

Duration of time charge allowed to pass = T = 49.0 min = 49.0 × 60 seconds

1 min = 60 seconds

Size of current : I

I=\frac{Q}{T}=\frac{N\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}

=\frac{n\times N_A\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}

I=\frac{0.550 mol\times 6.022\times 10^{23} mol^{-1}\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}=18.047 A\approx 18.0 A

18.0 Ampere is the size of electric current that must flow.

3 0
3 years ago
In a chemical reaction, one of the chemical reactants is present in______ to make sure that the limiting reagent is completely c
Nata [24]

Answer: excess

Explanation: i guesses and got the answer right

8 0
3 years ago
Suppose that you measured out 3.50g of Na2SO4. how many moles would you have? (show work)​
Ket [755]

Answer:

Na2SO4 is equal to 0.0070401642780093 mole.

Explanation:

5 0
3 years ago
A solution contains 0.60 mg/ml mn2+. what minimum mass of kio4 must me added to 5.00 ml of the solution in order to completely o
azamat
The given solution of Mn²⁺ is 0.60 mg/mL.
Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg

Molar mass of Mn = 54.9 g/mol
Hence, moles of Mn²⁺ = 3 x 10⁻³ g / 54.9 g/mol = 5.46 x 10⁻⁵ mol

The balanced equation for the reaction is,
2Mn²⁺ + 5KIO₄ + 3H₂O → 2MnO₄⁻ + 5KIO₃ + 6H⁺

The stoichiometric ratio between Mn²⁺ and KIO₄ is 2 : 5

Hence, moles of KIO₄ reacted = 5.46 x 10⁻⁵ mol x (5 / 2)
                                                 = 13.65 x 10⁻⁵ mol
Molar mass of KIO₄ = 230 g/mol

Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
                                               = 0.031395 g
                                               = 31.395 mg
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5 0
3 years ago
Which explains why krypton was most likely given this name
slavikrds [6]

Answer:

Krypton is named after the Greek word that means "secret." Which explains why krypton was most likely given this name? Krypton is a noble gas, so it was difficult for chemists to find it, as though it was a secret.

Explanation:

5 0
3 years ago
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