Which is larger? C atom or F atom? Explain why.

Describe how Gregor mendel's methods helped endure the accuracy of his results

aliina [53]

Not sure good luck on finding someone too help you

7 0

3 years ago

What do you think could be happening to the sand eels

Alex777 [14]

Answer:

is there like a passage to answer this question?

Explanation:

7 0

3 years ago

Read 2 more answers

Which situation is an example of heat transfer by radiation?

Artist 52 [7]

Answer is:
The sun's energy is transferred through the vacuum of space to Earth

5 0

3 years ago

Read 2 more answers

Consider the following balanced reaction. How many grams of water are required to form 75.9 g of HNO3? Assume that there is exce

adoni [48]

Answer:

The answer is "10.84 g".

Explanation:

The formula for calculating the number for moles:

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

In the given acid nitric:

Owing to the nitric acid mass = $75.9 g$

Nitric acid molar weight $= 63\ \frac{g}{mol}$

If they put values above the formula, they receive:

$\text{moles in nitric acid} = \frac{75.9}{63}$

$=1.204 \ mol$

In the given chemical equation:

$3 NO_2 \ (g) + H_2O \ (l) \longroghtarrow 2 HNO_3 \ (aq) + NO\ (g)$

In this reaction, 2 mols of nitric acid are produced by 1 mole of water.

So, 1.204 moles of nitric acid will be produced:

$= \frac{1}{2} \times 1.204$

$=0.602\ \ \text{mol of water}.$

We are now using Equation 1 in determining the quantity of water:

Water moles $= 0.602\ mol$

Water weight molar $= 18.02 \ \frac{g}{mol}$

$\to 0.602 = \frac{\text{mass of mols}}{ 18.02}\\\\\to \text{mass of mols} = 0.602 \times 18.02\\\\$

$=10.84 \ g$

8 0

3 years ago

Find the pH of a 0.010 M HNO2 solution.

lidiya [134]

Data:
Molar Mass of HNO2
H = 1*1 = 1 amu
N = 1*14 = 14 amu
O = 3*16 = 48 amu
------------------------
Molar Mass of HNO2 = 1 + 14 + 48 = 63 g/mol

M (molarity) = 0.010 M (Mol/L)

Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.010 M (Mol/L)
Use: Ka (ionization constant) = $5.0*10^{-4}$
$\alpha^2 (degree\:of\:ionization) = ?$

$Ka = M * \alpha^2$
$5.0*10^{-4} = 0.010* \alpha^2$
$0.010\alpha^2 = 5.0*10^{-4}$
$\alpha^2 = \frac{5.0*10^{-4}}{0.010}$
$\alpha^2\approx500*10^{-4}$

$\alpha\approx\sqrt{500*10^{-4}}$
$\alpha \approx 2.23*10^{-3}$

Now, we will calculate the amount of Hydronium [H3O+] in nitrous acid (HNO2), multiply the acid molarity by the degree of ionization, we will have:

$[ H_{3} O^+] = M* \alpha$
$[ H_{3} O^+] = 0.010* 2.23*10^{-3}$
$[ H_{3} O^+] \approx 0.0223*10^{-3}$
$[ H_{3} O^+] \approx 2.23*10^{-5} \:mol/L$

And finally, we will use the data found and put in the logarithmic equation of the PH, thus:

Data:
log10(2.23) ≈ 0.34
pH = ?
$[ H_{3} O^+] = 2.23*10^{-5}$

Formula:
$pH = - log[H_{3} O^+]$

Solving:
$pH = - log[H_{3} O^+]$
$pH = -log2.23*10^{-5}$
$pH = 5 - log2.23$
$pH = 5 - 0.34$
$\boxed{\boxed{pH = 4.66}}\end{array}}\qquad\quad\checkmark$

Note:. The pH <7, then we have an acidic solution.

6 0

3 years ago