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bagirrra123 [75]
3 years ago
6

Give the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by the element. rb

Chemistry
1 answer:
devlian [24]3 years ago
3 0

Actually Rb or Rubidium in zero state has the following electron configuration:

<span>1s22s2</span><span>2p6</span><span>3s2</span><span>3p63d10</span><span>4s2</span><span>4p65s1</span>

 

However we can see that the ion has a 1 positive charge, which means that it lacks 1 electron, therefore the answer from the choices is:

<span>d. rb+: 1s22s22p63s23p64s23d104p6</span>
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Question 5 (1 point)
Marianna [84]

Answer:

P=12.16 atm

Explanation:

Using the formula of ideal gas law:

PV = nRT

P= nRT/V

 n= number of moles

 R= Avogadro constant = 0.0821

 T= Temperature in K => ºC + 273.15 K

P= (1.50 moles)(0.0821)( 296.15 K)/ 3.00L

P= 12.15

7 0
3 years ago
P= f/a solve it for f
Bess [88]

Answer:

Force = Pressure × Area

Explanation:

Easy, just invert the equation. Transpose the force variable over to the left of the equals sign, and transpose the pressure variable back to the right side.

7 0
2 years ago
A laboratory procedure calls for making 590.0 mL of a 1.1 M KNO3 solution. How much KNO3 in grams is needed?
AveGali [126]
590 mL = 590 cm³= 0,59 dm³

C = n/V
n = 1,1M × 0,59 dm³
n = 0,649 mol
_____________________________

M KNO₃ = 39g+14g+16g×3 = 101 g/mol

1 mole -------- 101g
0,649 --------- X
X = 101×0,649
X = 65,549g KNO₃

:)
8 0
3 years ago
6. How many moles of water would require 92.048 kJ of heat to raise its temperature from 34.0 °C to 100.0 °C? (3 marks)​
scoray [572]

Taking into account the definition of calorimetry, 0.0185 moles of water are required.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

<h3>Mass of water required</h3>

In this case, you know:

  • Heat= 92.048 kJ
  • Mass of water = ?
  • Initial temperature of water= 34 ºC
  • Final temperature of water= 100 ºC
  • Specific heat of water = 4.186 \frac{J}{gC}

Replacing in the expression to calculate heat exchanges:

92.048 kJ = 4.186 \frac{J}{gC}× m× (100 °C -34 °C)

92.048 kJ = 4.186 \frac{J}{gC}× m× 66 °C

m= 92.048 kJ ÷ (4.186 \frac{J}{gC}× 66 °C)

<u><em>m= 0.333 grams</em></u>

<h3>Moles of water required</h3>

Being the molar mass of water 18 \frac{g}{mole}, that is, the amount of mass that a substance contains in one mole, the moles of water required can be calculated as:

amount of moles=0.333 gramsx\frac{1 mole}{18 grams}

<u><em>amount of moles= 0.0185 moles</em></u>

Finally, 0.0185 moles of water are required.

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8 0
2 years ago
What is the average yearly rate of change of carbon-14 during the first 5000 years?
erica [24]

Answer:

The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year

Explanation:

Given that the mass of the carbon 14 at the start = 5 gram

At the end of 5,000 years we will have;

A = A_0 \times e^{-\lambda \times t}

Where

A = The amount of carbon 14 left

A₀ = The starting amount of carbon 14

e = Constant = 2.71828

T_{1/2} = The half life

\lambda = 0.693/T_{1/2}

t = The time elapsed = 5000 years

λ = 0.693/T_{1/2} = 0.693/5730 = 0.0001209424

Therefore;

A = 5 × e^(-0.0001209424×5000) = 2.7312 grams

Therefore, the amount of carbon 14 decayed in the 5000 years is the difference in mass between the starting amount and the amount left

The amount of carbon 14 decayed = 5 - 2.7312 = 2.2688 grams

The average yearly rate of change of carbon-14 during the first 5000 years  is therefore;

2.2688 grams/(5000 years) = 0.0004538 grams per year

The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year.

5 0
3 years ago
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