CHEMISTRY COLLEGE

How many electrons does a Bromine (Br) atom have?(Given Information)
Atomic Number : 35
Neutrons: 45
Charge; -1

Answers

Answer 1
Answer: 35 yooooooooo...........

Related Questions

Given the following equation: 8 Fe + S8 ---> 8 FeSHow many moles of iron are needed to react with 16.0 moles of sulfur? A FULL STEP EXPLANATION WILL BE APPRECIATED
Group/family 18 on the periodic table is calledO alkaline earth metals Onoble gases O halogens O transitional metals
Pharmacists sometimes measure medicines in the unit of "grains," where 1 gr = 65 mg. The label on a bottle of aspirin reads "Aspirin, 5 gr." How many mg of aspirin are in a tablet containing 5 gr?
A chemical equation is balanced when thesubscripts of the reactants equal the subscripts of the products,coefficients of the reactants equal the coefficients of the products,products and reactants are the same chemicals,same number of each kind of atom appears in the reactants and in the products,
What is the polarity of a Na-Br bond?nonpolar covalent definitely polar covalent likely ionic slightly polar covalent slightly ionic

4.What volume of 0.120 M HNO3(aq) is needed to
completely neutralize 150.0 milliliters of 0.100 M
NaOH(aq)?
A. 62.5 mL
B. 125 ml
C.
180. mL
D. 360. mL
​

Answers

Answer:

B) 125 mL

Explanation:

M1V1=M2V2

(0.120M)(x)=(150.0 mL)(0.100M)

x= 125 mL

*Text me at 561-400-5105 for private tutoring if interested: I can do homework, labs, and other assignments :)

How would you explain the path of alpha particles in relation to each subatomic particle in the plum pudding model?

Answers

In the plum pudding model, alpha particles were expected to pass through the foil at different angles and some of them reflected. They realized that there are positive charge atom because it repels. So the plum pudding was replaced by the nuclear model of atom.

a student wants to remove the salt from a mixture of sand and salt in order to get only person he adds water to the mixture why it's the step help

Answers

You would add water to the mixture to dissolve the salt. When the salt is dissolved, you have your sand alone.
Water dissolves the salt and pour out the water let it evaporate and you have separate salt and water

100.0 g of liquid copper (molar mass 63.546 g/mol; melting point 1358 K; density 8.02 g/mL) is placed in a rigid container of volume 10.0 L at temperature 1508 K. The container is placed in an evacuated chamber and a small hole of area 3.23 mm2 is made in the upper container wall. After 2.00 hours, the mass of copper in the container has decreased by 1 0.0168 g. Assuming the mass loss is due to effusion, calculate the vapor pressure of liquid copper at 1508 K. Hint: because the liquid constantly evaporates, the pressure inside the container is constant

Answers

Answer:

8.912x10^-18

Explanation:

-dn/dt = pANa/2piMRT

100 g = initial copper

Number of moles = 100/63.546

= 1.5736

Mass of copper left = 100-10.0168

= 89.9832

Moles = 89.9832/63.546

= 1.4160

dn = 1.4160-1.5736

= -0.1576

dt = 2 hrs

A = 3.23mm² = 3.23x10^-6

M = 63.546

T = 0.0821

T = 1508k

Na = 6.023x10²³

When we insert all these into the formula above

We get

P = 8.912x10^-18atm

Consider the following reaction: Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)If you react an excess of Pb(NO3)2with 26.3 g of NaCl, and you isolate 52.1 g of PbCl2, what is your percent yield?

Answers

Answer:

\large \boxed{84.7 \, \%}

Explanation:

Mᵣ:                          58.44      278.11

           Pb(NO₃)₂ + 2NaCl ⟶ PbCl₂ + 2NaNO₃

m/g:                         26.3

1. Moles of NaCl

\text{Moles of NaCl} = \text{26.3 g NaCl} * \frac{\text{1 mol NaCl}}{\text{58.44 g NaCl}} = \text{0.4505 mol NaCl}

(b) Moles of PbCl₂

\text{Moles of PbCl${_2}$} = \text{0.4505 mol NaCl} * \frac{\text{1 mol PbCl${_2}$}}{\text{2 mol NaCl}} = \text{0.2253 mol PbCl${_2}$}

(c) Theoretical yield of PbCl₂

\text{Mass of PbCl${_2}$} = \text{0.2253 mol PbCl${_2}$} * \frac{\text{278.11 g PbCl${_2}$}}{\text{1 mol PbCl${_2}$}} = \text{61.52 g PbCl${_2}$}

(d) Percent yield

\text{Percent yield} = \frac{\text{ actual yield}}{\text{ theoretical yield}} * 100 \,\% = \frac{\text{52.1 g}}{\text{61.52 g}} * 100 \, \% = \mathbf{84.7 \,\%}\n\n\text{The percent yield is $\large \boxed{\mathbf{84.7 \, \% }}$}

In an attempt to prepare n−propylbenzene, a chemist alkylated benzene with 1−chloropropane and aluminum chloride. However, two isomeric hydrocarbons were obtained in a ratio of 2:1, the desired n−propylbenzene being the minor component. What do you think was the major product? How did it arise?

Answers

Answer:

2-Phenylpropane (Cumene)

Explanation:

Famous Friedel Craft Alkylation.

Aluminum Chloride grasped the 1-chloropropane forming an intermediate product composed of Aluminum tetrachloride and n-propylcation. It is well understood that primary carbocations are unstable and therefore undergo hydrogen shifting to attain stability. n-Propylcarbocation undergone hydrogen shifting, forming isopropylcarbocation which reacted with benzene forming 2-Phenylpropane as the major product and HCl as a byproduct.

AlCl3 + CH3CH2CH2Cl --> AlCl4-   +   CH3CH2CH2+

CH3CH2CH2+  ---> CH3CH(+)CH3

C6H6 + CH3CH(+)CH3 ---> C6H5CH(CH3)2   + H+

AlCl4- + H+ ---> HCl + AlCl3

Answer:

From the given problem statement,he was attempting to prepare n−propylbenzene by alkylation benzene with 1−chloropropane and aluminum chloride,but 1-propyle benze was a major product in result.
Other Questions
Question 1Ernest Rutherford completed his famous gold foil experiment in 1911. In this experiment, alpha particles were fired at a thin sheet of gold foil. He observed that most of the alpha particles passed straight through the gold foil unimpeded, but a small number of alpha particles were deflected. Which of the following conclusions about atomic structure were made from Rutherford’s gold foil experiment?
What is an acronym for the rainbow colors?I just want to know if anybody else knows this.
If a molecule diffusing through extracellular fluid travels 1 mm in 1 sec, how long will it take that molecule to diffuse 1 cm
How many moles of CO2 form when 4.0 mol of ethane, C2H6, react?2C2H6 + 702 + 4CO2 + 6H20A. 8 molB. 2 molC. 6 molD. 4 mol