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4 years ago

What two types of elements make up an Ionic bond ?

1 answer:

3 0

Answer it occurs between a metal and non metal eg sodium[Na]metal and chloride[Cl]a non metal form an ionic bond to make NaCl

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Can someone please help me please dont scroll

Nostrana [21]

Quantitative is anything that is relating to a number value. Think “quantity”

Example:

1. There are 5 ml of water
2. There are 3 beakers
3. It takes 30 minutes for the reaction

Qualitative is about characteristics that can be described

Example

1. The liquid is green
2. The solution turned into a gas
3. Bubbles were produced

4 0

3 years ago

A couple is planning to have a child. the father-to-be has six fingers, a dominant trait. His genotype is Ff. His wife has five

postnew [5]

I think it’s C.1/2 because the child could get five fingers like his mother or six fingers like his dad

4 0

4 years ago

4. Calculate the total heat required to convert 23.6 g of ice at -27 ºC to steam at 121 ºC. Hint: Calculation involves more than

DanielleElmas [232]

Answer:

filedownload.ashx heres a pdf

Explanation:

8 0

3 years ago

At a particular temperature a 2.00-L flask at equilibrium contains 2.80 ✕ 10-4 mol N2, 2.50 ✕ 10-5 mol O2, and 2.00 ✕ 10-2 mol N

zhenek [66]

Answer : The value of equilibrium constant (K) is, $5.71\times 10^4$

Explanation :

First we have to calculate the concentration of $N_2,O_2\text{ and }N_2O$

$\text{Concentration of }N_2=\frac{\text{Moles of }N_2}{\text{Volume of solution}}=\frac{2.80\times 10^{-4}mol}{2.00L}=1.4\times 10^{-4}M$

and,

$\text{Concentration of }O_2=\frac{\text{Moles of }O_2}{\text{Volume of solution}}=\frac{2.50\times 10^{-5}mol}{2.00L}=1.25\times 10^{-5}M$

and,

$\text{Concentration of }N_2O=\frac{\text{Moles of }N_2O}{\text{Volume of solution}}=\frac{2.00\times 10^{-2}mol}{2.00L}=1.00\times 10^{-2}M$

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

$N_2(g)+O_2(g)\rightarrow 2N_2O(g)$

The expression for equilibrium constant is:

$K=\frac{[N_2O]^2}{[N_2][O_2]}$

Now put all the given values in this expression, we get:

$K=\frac{(1.00\times 10^{-2})^2}{(1.4\times 10^{-4})\times (1.25\times 10^{-5})}$

$K=5.71\times 10^4$

Thus, the value of equilibrium constant (K) is, $5.71\times 10^4$

7 0

4 years ago

Calculate the number of pounds of CO2 released into the atmosphere when a 15.0-gallon tank of gasoline is burned in an automobil

masha68 [24]

<h3>Answer:</h3>

266.997 pounds

<h3>Explanation:</h3>

The balanced equation for the combustion of octane is;

2C₈H₁₈(l) + 25O₂(g) → 16CO₂(g) + 18H₂O(g)

Assuming that gasoline is purely octane:

Amount of gasoline available is 15.0 gallon

We are required to calculate the amount of CO₂ released to the atmosphere;

This can be done in several steps;

Step 1: Mass of octane in grams

Amount of octane = 15.0 gallons

But, 1 gallon = 3785.41 ml

Therefore;

15 gallon is equivalent to;

= 15 ×3785.41 ml

= 56,781.15 ml

We are given the density of octane (0.692 g/ml) and thus we can calculate the mass of octane.

Mass = density × volume

= 0.692 g/ml × 56,781.15 ml

= 39,292.56 g

Step 2: Number of moles of octane

Number of moles = Mass/ Molar mass

Molar mass of octane = 114.23 g/mol

Number of moles of Octane = 39,292.56 g ÷ 114.23 g/mol

= 343.978 moles

Step 3: Moles of carbon dioxide released

From the equation;

2 moles of octane completely burns in air to yield 16 moles of CO₂

Therefore;

343.978 moles of octane produces;

= (343.978 moles/2) 16

= 2751.824 moles of CO₂

Step 4: Mass of CO₂ in pounds

1 mole of CO₂ has 44.01 g

Therefore;

2751.824 moles contains;

= 2751.824 moles × 44.01 g/mol

= 121,107.77 g

But, 1 pound = 453.592 g

Therefore;

Mass of CO₂ in pounds = 121,107.77 g ÷ 453.592 g

= 266.997 pounds

Therefore, the mass of CO₂ released to the atmosphere is 266.997 pounds

4 0

3 years ago