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3 years ago

How many valence electrons does Ne have

Chemistry

2 answers:

ollegr [7]3 years ago

6 0

Answer:

eight valence electrons

lidiya [134]3 years ago

5 0

Answer:

it has 8 valence electrons

You might be interested in

The percent yield of this reaction is consistently 92.0%. CH4(g) + 4 S(g) → CS2(g) + 2 H2S(g) How many grams of sulfur would be

dybincka [34]

Answer:

132.17 g

Explanation:

The reaction given , in the question is -

CH₄ (g ) + 4 S ( g ) ---> CS₂ ( g ) + 2H₂S ( g )

From the reaction , 4 mole of S is required for the production of 1 mole of CS₂ .

since ,

Moles of CS₂ = given mass of CS₂ / Molecular weight of CS₂

Since ,

the Molecular weight of CS₂ = 76

Given , mass of CS₂ = 72.57 g

Moles of CS₂ = 72.57 / 76 = 0.95 mol

Since ,

The yield is 92.0 % .

Moles of S required = 4 * 0.95 mol / 0.92 = 4.13 moles

Mass of S required = 4.13 * 32 = 132.17 g .

6 0

3 years ago

The solubility of acetanilide is 12.8 g in 100 mL of ethanol at 0 ∘C, and 46.4 g in 100 mL of ethanol at 60 ∘C. What is the maxi

weqwewe [10]

Answer: 72.41% and 26.90% respectively.

Explanation:

At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.

We can calculate recovery as:

$\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{33.6\ g}{46.4\ g}*100=72.41\%$

So the answer to the first question is 72.41%.

For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.

$\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{46\ mg}{171\ mg}*100=26.90\%$

So the answer to the second question is 26.90%.

3 0

3 years ago

what is the concentration of a NaCl solution, in Molarity, if you add 59.76 g of NaCl into 270 mL of H2O

enot [183]

Hey there!

Molar mass NaCl = 58.44 g/mol

Number of moles

n = mass of solute / molar mass

n = 59.76 / 58.44

n = 1.0225 moles of NaCl

Volume in liters:

270 mL / 1000 => 0.27 L

Therefore:

M = number of moles / volume ( L )

M = 1.0225 / 0.27

= 3.78 M

Hope that helps!

7 0

3 years ago

The maximum amounts of lead and copper allowed in drinking water are 0.015 mg/kg for lead and 1.3 mg/kg for copper. Tell the max

GuDViN [60]

Answer:

The maximum amount of copper allowed in 100 g of water is 0.00013 g

Explanation:

To find the maximum amount of copper (in grams) allowed in 100 g of water use the maximum amount ratio (1.3 mg / kg) and set a proportion with the unknown amount of copper (x) and the amount of water (100 g):

First, convert 100 g of water to kg: 100 g × 1 kg / 1000 g = 0.1 kg.

Now, set the proportion:

1.3 mg Cu / 1 Kg H₂O = x / 0.1 kg H₂O

Solve for x:

x = 0.1 kg H₂O × 1.3 mg Cu / 1 kg H₂O = 0.13 mg Cu

Convert mg to grams:

0.13 mg × 1 g / 1,000 mg = 0.00013 g

Answer: 0.00013 g of copper.

6 0

3 years ago

For the gas phase decomposition of 1-bromopropane, CH3CH2CH2BrCH3CH=CH2 + HBr the rate constant at 622 K is 6.43×10-4 /s and the

ladessa [460]

Answer is: activation energy of this reaction is 212,01975 kJ/mol.
Arrhenius equation: ln(k₁/k₂) = Ea/R (1/T₂ - 1/T₁).
k₁ = 0,000643 1/s.
k₂ = 0,00828 1/s.

T₁ = 622 K.

T₂ = 666 K.

R = 8,3145 J/Kmol.

1/T₁ = 1/622 K = 0,0016 1/K.
1/T₂ = 1/666 K = 0,0015 1/K.
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol · (-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol · (-0,0001 1/K).
Ea = 212019,75 J/mol = 212,01975 kJ/mol.

4 0

3 years ago