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Tcecarenko [31]
3 years ago
8

If 250 g of calcium chlorate decomposes into calcium chloride and oxygen what mass of oxygen would be produced

Chemistry
1 answer:
andrey2020 [161]3 years ago
4 0

Answer:

m O2 = 116,16 g

Explanation:

Ca(ClO3)2 ------------> CaCl2 + 3O2

n Ca(ClO3)2 = 250/207 = 1,21 mol

n O2 = 3n Ca(ClO3)2 = 1,21 * 3 = 3,63 mol

m O2 = 3,63 * 32 = 116,16 g

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DiKsa [7]

Answer:

75000 Hz

Explanation:

f = V / λ (f= frequency, v=velocity of wave, lambda= wavelength)

alternatively, f = c / λ (f= frequency, c= speed of light- 3.00x10^8 m/s, lambda= wavelength)

f= [3.00x10^8 m/s]/[4000 m]

=75000 Hz

3 0
3 years ago
How many grams of Lioh are in 1.7 moles please help
natali 33 [55]

Answer:

23.94834 gram

Explanation:

4 0
3 years ago
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Before tackling this problem, be sure you know how to find the antilog of a number using a scientific calculator.
dybincka [34]
<h2>Question:- </h2>

A solution has a pH of 5.4, the determination of [H+].

<h2>Given :- </h2>
  1. pH:- 5.4
  2. pH = - log[H+]

<h2>To find :- concentration of H+</h2>

<h2>Answer:- Antilog(-5.4) or 4× 10-⁶</h2>

<h2>Explanation:- </h2><h3>Formula:- pH = -log H+ </h3>

Take negative to other side

-pH = log H+

multiple Antilog on both side

(Antilog and log cancel each other )

Antilog (-pH) = [ H+ ]

New Formula :- Antilog (-pH) = [+H]

Now put the values of pH in new formula

Antilog (-5.4) = [+H]

we can write -5.4 as (-6+0.6) just to solve Antilog

Antilog ( -6+0.6 ) = [+H]

Antilog (-6) × Antilog (0.6) = [+H]

Antilog (-6)  = {10}^{ - 6} ,  \\ Antilog (0.6)  = 4

put the value in equation

{10}^{ - 6}   \times 4 = [H+] \\ 4 \times   {10}^{ - 6}  = [H+]

7 0
2 years ago
Read 2 more answers
10. Which is an example of mechanical energy?*
gtnhenbr [62]

Answer: A golfer hitting a golf ball.

Explanation:

The atomic particles move more in this option than the others.

6 0
3 years ago
A buffer solution is composed of 4.00 4.00 mol of acid and 3.25 3.25 mol of the conjugate base. If the p K a pKa of the acid is
Reika [66]

<u>Answer:</u> The pH of the buffer is 4.61

<u>Explanation:</u>

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{conjuagate base}]}{[\text{acid}]})

We are given:

pK_a = negative logarithm of acid dissociation constant of weak acid = 4.70

[\text{conjuagate base}]} = moles of conjugate base = 3.25 moles

[\text{acid}]  = Moles of acid = 4.00 moles

pH = ?

Putting values in above equation, we get:

pH=4.70+\log(\frac{3.25}{4.00})\\\\pH=4.61

Hence, the pH of the buffer is 4.61

8 0
3 years ago
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