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2 years ago

Is krypton a anion or cation?

Chemistry

2 answers:

Zarrin [17]2 years ago

7 0

Answer:

krypton is a cation

magnesium is also a cation

aluminium is a anion

and , silicon is cation

cluponka [151]2 years ago

7 0

Answer:

• Krypton: cation
• Magnesium: cation
• Aluminum: anion
• Silicone: cation

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What are 5 responsibilities a citizen has in taking part in public affairs

zimovet [89]

1. To vote
2. To abide by laws
3. To contribute to society positively
4. To be kind to people around you
<span>5. To pass on the top four to your children and neighbors</span>

3 0

3 years ago

What causes atoms to form covalent bonds?

navik [9.2K]

Answer:

to attain stability atoms form covalent bonds

7 0

2 years ago

How much potassium chlorate would dissolve in 200g water at 20°C?

DedPeter [7]

Answer:

14 solubility in 200 gram of water at 20 c

Explanation:

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6 0

3 years ago

A sample of 87.6 g of carbon is reacted with 136 g of

Vadim26 [7]

Answer:

A. fluorine, 1.79 moles

Explanation:

Given parameters:

Mass of carbon = 87.7g

Mass of fluorine gas = 136g

Unknown:

The limiting reactant and the maximum amount of moles of carbon tetrafluoride that can be produced = ?

Solution:

Equation of the reaction:

C + 2F₂ → CF₄

let us find the number of the moles the given species;

Number of moles = $\frac{mass}{molar mass}$

C; molar mass = 12;

Number of moles = $\frac{87.7}{12}$ = 7.31moles

F; molar mass = 2(19) = 38g/mol

Number of moles = $\frac{136}{38}$ = 3.58moles

So;

From the give reaction:

1 mole of C requires 2 moles of F₂

7.31 moles of C will then require 2 x 7.31 moles of F₂ = 14.62moles

But we have 3.58 moles of the F₂;

Therefore, the reactant in short supply is F₂ and it is the limiting reactant;

So;

2 moles of F₂ will produce mole of CF₄

3.58 moles of F₂ will then produce $\frac{3.58}{2}$ = 1.79moles of CF₄

6 0

3 years ago

1) A light bulb takes in 30 of energy per second. It transfers 3j as use

natta225 [31]

Answer:

$\boxed{\text{10 \%}}$

Explanation:

The formula for efficiency is

$\begin{array}{rcl}\text{Efficiency} & = & \dfrac{\text{useful energy out}}{\text{energy in}} \times 100 \,\% \\\\\eta & = & \dfrac{w_{\text{out}}}{w_{\text{in}} } \times 100 \,\%\\\end{array}$

Data:

Useful energy = 3 J

Energy input = 30 J

Calculation:

$\begin{array}{rcl}\eta & = & \dfrac{\text{3 J}}{\text{30 J}} \times 100 \,\%\\\\\eta & = & 10 \, \%\\\end{array}\\\text{ The efficiency is }\boxed{\textbf{10 \%}}$

8 0

3 years ago