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N76 [4]
3 years ago
11

Can someone help me solve this? Thanks!

Chemistry
1 answer:
IrinaVladis [17]3 years ago
5 0

Answer:

yea

Explanation:

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Please help on this one?
Bezzdna [24]

Answer:

\text{C. } _{36}^{85}\text{Kr}

Explanation:

Your nuclear equation is  

_{35}^{85}\text{Br} \longrightarrow \, _{-1}^{0}\text{e} +\, _{x}^{y}\text{X}

The main point to remember in balancing nuclear equations is that

  • the sum of the superscripts and must be the same on each side of the equation.
  • the sum of the subscripts must be the same on each side of the equation.  

Then  

85 = 0 + y, so y = 85 - 0 = 0  

35 = -1 + x, so x = 35 + 1 = 36

The nucleus with atomic number 36 and atomic mass 85 is krypton-85.  

The nuclear equation becomes  

_{35}^{85}\text{Br} \longrightarrow \, _{-1}^{0}\text{e} + \, _{36}^{85}\text{Kr}

4 0
3 years ago
Read 2 more answers
Concentrations-
Arisa [49]
Moles of HCl is 3.47mol

Work shown on photo

6 0
3 years ago
Assuming the densities are the same as water, calculate the concentration in millimolar (mM) for all three reactant solutions, t
Marina86 [1]

Answer:

foatation

Explanation:

7 0
3 years ago
Cesium has a radius of 272 pm and crystallizes in a body-centered cubic structure. What is the edge length of the unit cell
olchik [2.2K]

Answer: Edge length of the unit cell = 628pm

Explanation: For a body centred cubic structured system, the relationship between the edge length of the unit cell and radius of the atoms in the structure is

Edge length of Unit cell (a) = (4R)/(√3)

R = 272pm = (272 × (10^-12))m = (2.72 × (10^-10))m

a = (4 × (2.72 × (10^-10)))/(√3)

a = (6.28157 × (10^-10))m = 628pm

4 0
3 years ago
A machine shop worker records the mass of an aluminum cube as 176 g. If one side of the cube measures 4 cm, what is the density
forsale [732]

Answer:

Explanation:

So we take the given quotient:

ρ=176⋅g4⋅cm×4⋅cm×4⋅cm = 176⋅g64⋅cm3 = ??g⋅cm−3.

Would the cube float or sink in water? Why

6 0
3 years ago
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