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Zinaida [17]
3 years ago
14

Electron configurations of bromine short and long

Chemistry
1 answer:
nekit [7.7K]3 years ago
6 0

Explanation:

short electronic configuration of bromine

  • [Ar] 3d¹⁰ 4s² 4p⁵

long electronic configuration of bromine

  • 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁵
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genetics and reproduction is all about dna.

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What are atoms?
liq [111]

Answer:

4. particles of matter

8 0
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Read 2 more answers
You add 100.0 g of water at 52.0 °C to 100.0 g of ice at 0.00 °C. Some of the ice melts and cools the water to 0.00 °C. When the
lara31 [8.8K]

Answer:

m_{ice} = 65.336\,g

Explanation:

Accoding to the First Law of Thermodynamics, the heat released by the water melts a portion of ice. That is to say:

Q_{water} = Q_{ice}

(100\,g)\cdot \left(4.184\,\frac{J}{kg\cdot ^{\textdegree}C}\right)\cdot (52\,^{\textdegree}C - 0\,^{\textdegree}C) = m_{ice}\cdot \left(333\,\frac{J}{g} \right)

The amount of ice that is melt is:

m_{ice} = 65.336\,g

5 0
3 years ago
A hot lump of 30.5 g of iron at an initial temperature of 52.7 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to r
slava [35]

Answer:

26.7°C

Explanation:

Using the formula; Q = m × c × ΔT

Where; Q = amount of heat

m = mass

c = specific heat

ΔT = change in temperature

In this question involving iron placed into water, the Qwater = Qiron

For water; m= 50g, c = 4.18 J/g°C, Initial temp= 25°C, final temp=?

For iron; m = 30.5g, c = 0.449J/g°C, Initial temp= 52.7°C, final temp=?

Qwater = -(Qiron)

m × c × ΔT (water) =- {m × c × ΔT (iron)}

50 × 4.18 × (T - 25) = - {30.5 × 0.449 × (T - 52.7)}

209 (T - 25) = - {13.6945 (T - 52.7)}

209T - 5225 = -13.6945T + 721.7

209T + 13.6945T = 5225 + 721.7

222.6945T = 5946.7

T = 5946.7/222.6945

T = 26.7

Hence, the final temperature of water and iron is 26.7°C

8 0
3 years ago
An excess of mg(s) is added to 100.ml of 0.400 m hcl. at 0c and 1 atm pressure, what volume of h 2 (g) can be obtained?
ra1l [238]
The balanced equation for the reaction between Mg and HCl is as follows
Mg + 2HCl --> MgCl₂ + H₂
stoichiometry of HCl to H₂ is 2:1

number of HCl moles reacted - 0.400 mol/L x 0.100 L = 0.04 mol of HCl
since Mg is in excess HCl is the limiting reactant 
number of H₂ moles formed - 0.04/2 = 0.02 mol of H₂

we can use ideal gas law equation to find the volume of H₂
PV = nRT 
where 
P - pressure - 1 atm x 101 325 Pa/atm = 101 325 Pa
V - volume
n - number of moles - 0.02 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 0 °C + 273 = 273 K
substituting these values in the equation 

101 325 Pa x V = 0.02 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K
V = 448 x 10⁻⁶ m³
V = 448 mL 
therefore answer is 
c. 448 mL 
5 0
3 years ago
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