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masya89 [10]
3 years ago
5

How many neutrons does 85Rb have?

Chemistry
1 answer:
Leona [35]3 years ago
8 0

Answer:

Rb has two isotopes - one having 48 neutrons and the other having 50 neutrons. The atomic mass number of an isotope is the sum of the protons plus neutrons in the nucleus, so the two stable isotopes of Rb are designated 85Rb and 87Rb (their atomic mass numbers).

HOPE THIS HELPED!!!!!!!!!!!!!!! XDDDDDDDDDD

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How much volume would a 834.01g pile of sugar have given that it has a density of 1.59g/mL?
boyakko [2]

Answer:

v = 534.5mL

m = 597.15g

Density = 9.23g/mL

Density = 9.125g/mL

Explanation:

Density = mass/ volume

For the first question

Density = 1.59g/mL

Mass = 834.01g

Volume = ?

Using the above formula we have 1.59 = 834.01/v

v = 834.01/1.59

v = 534.5mL

For the second question

Density =0.9167g/mL

Volume = 651.41mL

Mass =?

Using the above formula we have

0.9167 =m/651.41

Cross multiply

m = 0.9167 x 651.41

m = 597.15g

For the third question

Mass =803.44g

Volume=87.03mL

Density =?

Density = 803.44/87.03

= 9.23g/mL

For the fourth

Density = 56.85/6.23

= 9.125g/mL

7 0
3 years ago
The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions
mamaluj [8]

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

3 0
3 years ago
Does the identity of gas matter when predicting its behavior why or why not?
just olya [345]

Answer: The gas phase is unique among the three states of matter in that there are some simple models we can use to predict the physical behavior of all gases—independent of their identities. We cannot do this for the solid and liquid states. ... Gas particles do not experience any force of attraction or repulsion with each other.

Explanation:

4 0
3 years ago
Balancing chemical equations
kolbaska11 [484]

Answer:

2B2 + 3O2 → 2B2O3

Explanation:

Balance The Equation: B2 + O2 = B2O3

1. Label Each Compound With a Variable

  aB2 + bO2 = cB2O3

2. Create a System of Equations, One Per Element

  B: 2a + 0b = 2c

  O: 0a + 2b = 3c

3. Solve For All Variables (using substitution, gauss elimination, or a calculator)

  a = 2

  b = 3

  c = 2

4. Substitute Coefficients and Verify Result

  2B2 + 3O2 = 2B2O3

      L R

  B: 4 4 ✔️

  O: 6 6 ✔️

hope this helps!

7 0
2 years ago
According to the second law of thermodynamics, energy tends to become more spread out. True or False?
chubhunter [2.5K]
The answer is true. According to the second law of thermodynamics, energy tends to become more spread out 
8 0
3 years ago
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