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3 years ago

Show how the total charge sodium oxide is zero

1 answer:

6 0

First, the symbol for sodium oxide is Na₂O

Each Na (sodium) has a charge of 1+, and each O has a charge of 2- :

Na₂¹⁺O²⁻

There are two Na's, however, and each one is 1+, however, so the Na₂ has a total charge of 2+. Because of this, the 2+ from the 2 Na's and the 2- from the O cancel each other out to make 0.

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How many isotopes does argon have?

Nostrana [21]

Argon has 24 known isotopes.

7 0

3 years ago

The map shows the average high temperatures in July for two cities in Texas.

alexira [117]

Answer:

Highest temperature which was recorded in Texas was 40 degrees.

Explanation:

Texas has variating weather. In summer Texas city turns to be very hot and during winters the city of Texas experience fall in temperature. There is very few rainfall in Texas due the climatic conditions.

8 0

2 years ago

The heat of vaporization of water at 100°c is 40.66 kj/mol. Calculate the quantity of heat that is absorbed/released when 9.00 g

timofeeve [1]

Answer:

20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.

Explanation:

Heat is being consumed during vaporization and heat is being released during condensation.

To vaporize 1 mol of water, 40.66 kJ of heat is being consumed.

Molar mass of water = 18.02 g/mol

Hence, to vaporize 18.02 g of water , 40.66 kJ of heat is being consumed.

So, to vaporize 9.00 g of water, $(\frac{40.66}{18.02}\times 9.00)kJ$ of heat or 20.3 kJ of heat is being consumed

As condensation is a reverse process of vaporization therefore 20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.

5 0

3 years ago

Which statement concerning the hydrated hydrogen ion is incorrect? (a) Although we often represent hydrogen ions as bare protons

Harrizon [31]

Answer:C. The value of n for H+(H2O)n can be calculated for almost all solutions.

Explanation:

An hydrate can be described as a substance that contains water or with an hydrogen bonded water molecule group.

The hydrate group doesn't necessarily have a fixed formula.

5 0

3 years ago

What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$

The equilibrium reaction for dissociation of $HNO_2$ (weak acid) is,

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha )$ .

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha )$ .

$4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}$

$4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2$

$0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0$

By solving the terms, we get

$\alpha=0.0533$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.150\times 0.0533=0.007995 M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.007995 M)$

$pH=2.097\approx 2.1$

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0

3 years ago