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RSB [31]
3 years ago
8

Why does a beta particle have an atomic number of -1

Chemistry
1 answer:
Fudgin [204]3 years ago
6 0
I cant entirely tell for now but an article on rodioactivity should solve the problem
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What best explains whether bromine (Br) or neon (Ne) is more likely to form a covalent bond?
solniwko [45]

Answer:

Bromine forms covalent bonds because it has seven valence electrons, but neon has eight valence electrons and already fulfills the octet rule. Neutral atoms coming together to share electrons.

Hoped this helped

4 0
4 years ago
4.2 g of 1,4-di-t-butyl-2,5-dimethoxybenzene (250.37 g/mol) were synthesized by reacting 10.4 mL of t-butyl alcohol (MW 74.12 g/
nalin [4]

Answer:

Percentage yield of 1,4-di-t-butyl-2,5-dimethoxybenzene is 41.40%.

Explanation:

Here, in the reaction sulfuric acid is playing the role of catalyst by donating its proton in initial stage of the reaction and in the end of the reaction the proton is returned back to sulfuric acid.

Mass = Density × Volume

Mass of t-butyl alcohol = 0.79 g/mL\times 10.4 mL=8.219 g

Moles of t-butyl alcohol  =\frac{8.219 g}{74.12 g/mol}=0.11084 mol

Moles of 1,4-dimethoxybenzene = \frac{5.6 g}{138.17 g/mol}=0.04052 mol

According to reaction 2 mol of  t-butyl alcohol reacts  with 1 mol of 1,4-dimethoxybenzene.

Then 0.11084 moles of t-butyl alcohol will react with :

\frac{1}{2}\times 0.11084 mole=0.05542 mol of 1,4-dimethoxybenzene.

This means that moles of 1,4-dimethoxybenzene are limited and moles of t-butyl alcohol are in excess.So, the moles of product will depend upon the moles of 1,4-dimethoxybenzene.

According top reaction 1 mol of 1,4-dimethoxybenzene gives 1 mol of  1,4-di-t-butyl-2,5-dimethoxybenzene.

Then 0.04052 moles of 1,4-di-t-butyl-2,5-dimethoxybenzene will give:

\frac{1}{1}\times 0.04052 mol= 0.04052 mol of 1,4-di-t-butyl-2,5-dimethoxybenzene.

Mass of 0.04052 moles of 1,4-di-t-butyl-2,5-dimethoxybenzene:

0.04052 mol × 250.37 g/mol = 10.144 g

Percentage yield:

\frac{Experimental}{Theoretical}\times 100

Percentage yield of 1,4-di-t-butyl-2,5-dimethoxybenzene:

Experimental yield = 4.2 g

Theoretical yield = 10.144 g

\frac{4.2 g}{10.144 g}\times 100=41.40\%

4 0
3 years ago
Question 2 (1 point)<br>At the center of the atom is the electron.<br>True<br>False​
Stells [14]

Answer:

False the electrons are on the outside of the atoms

Explanation:

5 0
4 years ago
100.0 mL of Ca(OH)2 solution is titrated with 5.00 x 10–2 M HBr. It requires 36.5 mL of the acid solution for neutralization. Wh
miskamm [114]

Answer:

The number of moles HBr = 0.001825

The concentration of Ca(OH)2 = 0.009125 M

Explanation:

Step 1: Data given

Volume of the Ca(OH)2 = 100.0 mL = 0.100 L

Molarity of HBr = 5.00 * 10^-2 M

Volume of HBR = 36.5 mL = 0.0365 L

Step 2: The balanced equation

Ca(OH)2 + 2HBr → CaBr2 + 2H2O

Step 3: Calculate molarity of Ca(OH) 2

b*Va* Ca = a * Vb*Cb

⇒with b = the coefficient of HBr = 2

⇒with Va = the volume of Ca(OH)2 = 0.100 L

⇒with ca = the concentration of Ca(OH)2 = TO BE DETERMINED

⇒with a = the coefficient of Ca(OH)2 = 1

⇒with Vb = the volume of HBr = 0.0365 L

⇒with Cb = the concentration of HBr = 5.00 * 10^-2 = 0.05 M

2 * 0.100 * Ca = 1 * 0.0365 * 0.05

Ca = (0.0365*0.05) / 0.200

Ca = 0.009125 M

Step 4: Calculate moles HBr

Moles HBr = concentration HBr * volume HBr

Moles HBr = 0.05 M * 0.0365 L

Moles HBr = 0.001825 moles

3 0
4 years ago
Manganese(iv) oxide reacts with aluminum to form elemental manganese and aluminum oxide: 3mno2+4al→3mn+2al2o3part awhat mass of
Oxana [17]
<span>12.4 g First, calculate the molar masses by looking up the atomic weights of all involved elements. Atomic weight manganese = 54.938044 Atomic weight oxygen = 15.999 Atomic weight aluminium = 26.981539 Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol Now determine the number of moles of MnO2 we have 30.0 g / 86.936044 g/mol = 0.345081265 mol Looking at the balanced equation 3MnO2+4Al→3Mn+2Al2O3 it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So 0.345081265 mol / 3 * 4 = 0.460108353 mol So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum. 0.460108353 mol * 26.981539 g/mol = 12.41443146 g Finally, round to 3 significant figures, giving 12.4 g</span>
7 0
4 years ago
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