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4 years ago

List two ways that the first ionization and second ionization of an atom are similar

1 answer:

6 0

The first ionization and second ionization of an atom are similar in following ways:
1. Both ionizations involve atom and energy
2. Both lose an electron.

The difference between first and second ionization of an atom is that both started and ended with different ions/atom.

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What does it takes to prove hypothesis false

olganol [36]

Answer:A hypothesis or model is called falsifiable if it is possible to conceive of an experimental observation that disproves the idea in question.

Explanation:That is, one of the possible outcomes of the designed experiment must be an answer, that if obtained, would disprove the hypothesis.

3 0

3 years ago

Anyone know! Please help asap ty

DiKsa [7]

Answer:

An earthquake happens by the child being born from fat mama

Explanation:

6 0

3 years ago

Non-electrolyte = A, 10.6 grams, dissolved in solvent B, 740 grams, the boiling point of the solution is higher than the boiling

Thepotemich [5.8K]

Answer:

2.1 °C/m

Explanation:

Hello, for this exercise, consider the formula:

$T_{solution}-T{solvent}=K_bm_solute$

Considering that the difference in the temperature is 0.284°C, and the given molality by:

$m_{solute}=\frac{10.6g\frac{1mol}{106g}}{740g*\frac{1kg}{1000g} } \\m_{solute}=0.135m$

Now, solving for $K_b$ , we get:

$K_b=\frac{0.284C}{0.135m}\\K_b=2.1 C/m$

Best regards.

8 0

3 years ago

Balance each equation by entering the correct coefficients.

natali 33 [55]

Answer:

5SiO2 + 2CaC2 --> 5Si + 2CaO + 4CO2

4NH3 + 5O2 --> 4NO + 6H2O

6 0

3 years ago

A science teacher has a supply of 20% saline solution and a supply of 50% saline solution. How much of each solution should the

lilavasa [31]

Answer:

Explanation:

$\text{Let}\\\\x-\text{a volume of}\ 20\%\ \text{saline solution}\\\\y-\text{a volume of }\ 50\%\ \text{saline solution}\\\\p\%=\dfrac{p}{100}\\\\20\%=\dfrac{20}{100}=\dfrac{2}{10}=0.2\\\\50\%=\dfrac{50}{100}=\dfrac{5}{10}=0.5\\\\0.2x-\text{a volume of salt in }\ 20\%\ \text{saline solution}\\0.5y-\text{a volume of salt in }\ 50\%\ \text{saline solution}\\\\60mL-\text{a volume of}\ 28\%\ \text{saline solution}\\\\28\%=\dfrac{28}{100}=0.28$

$0.28\cdot60mL=16.8mL-\text{a volume of salt in }\ 28\%\ \text{saline solution}\\\\\text{Equations}\\\\(1)\qquad x+y=60\\(2)\qquad0.2x+0.5y=16.8$

$\left\{\begin{array}{ccc}x+y=60\\0.2x+0.5y=168&\text{multiply both sides by 10}\end{array}\right\\\left\{\begin{array}{ccc}x+y=60&\text{multiply both sides by (-2)}\\2x+5y=168}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}-2x-2y=-120\\2x+5y=168}\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad3y=48\qquad\text{divide both sides by 3}\\.\qquad y=16$

$\text{Put the value of}\ y\ \text{to the first equation:}\\\\x+16=60\qquad\text{subtract 16 from both sides}\\x=44$

6 0

4 years ago