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2 years ago

Balance CH3CHO + O2 → CO2 + H2O

Chemistry

1 answer:

babunello [35]2 years ago

7 0

2 CH3CHO + 5 O2 = 4 CO2 + 4 H2O

reaction type = combustion

hope this helped :)

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Now that Snape and Dumbledore has taught you the finer points of hydration calculations they have a slightly more challenging pr

Mrrafil [7]

Answer:

The value of an integer x in the hydrate is 10.

Explanation:

$Molarity=\frac{Moles}{Volume(L)}$

Molarity of the solution = 0.0366 M

Volume of the solution = 5.00 L

Moles of hydrated sodium carbonate = n

$0.0366 M=\frac{n}{5.00 L}$

$n=0.0366 M\times 5 mol=0.183 mol$

Mass of hydrated sodium carbonate = n= 52.2 g

Molar mass of hydrated sodium carbonate = 106 g/mol+x18 g/mol

$n=\frac{\text{mass of Compound}}{\text{molar mass of compound}}$

$0.183 mol=\frac{52.2 g}{106 g/mol+x\times 18 g/mol}$

$106 g/mol+x\times 18 g/mol=\frac{52.2 g}{0.183 mol}$

Solving for x, we get:

x = 9.95 ≈ 10

The value of an integer x in the hydrate is 10.

6 0

3 years ago

If 35.0 mL of water in a graduated cylinder is displaced by 8.00 mL

sdas [7]

Answer:

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume}\\$

From the question

volume = final volume of water - initial volume of water

volume = 35 - 8 = 27 mL

We have

$density = \frac{14}{27} \\ = 0.518518...$

We have the final answer as

Hope this helps you

3 0

2 years ago

What is the pH if the [H+] concentration is 3 x10^-13

Shkiper50 [21]

Answer:

pH = 12.52

Explanation:

Given that,

The [H+] concentration is $3\times 10^{-13}$ .

We need to find its pH.

We know that, the definition of pH is as follows :

$pH=-log[H^+]$

Put all the values,

$pH=-log[3\times 10^{-13}]\\\\pH=12.52$

So, the pH is 12.52.

6 0

3 years ago

The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge,

Mila [183]

Answer:

1.33 Å

Explanation:

Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å

Also,

For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.

Thus,

$\frac {r^+}{r^-}=0.731$ .................1

Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.

Thus,

$r^++r^-=\frac {a}{2}$ ...................2

Given that:

$Cl^-\ (r^-) = 1.82\ \dot{A}$

To find,

$K^+\ (r^+) = ? \dot{A}$

Using 1 and 2 , we get:

$1.731\ r^+=0.731\times \frac {6.28}{2}$

<u>Size of the potassium ion = 1.33 Å</u>

4 0

3 years ago

Your body is much larger now than it was when you were a baby. What is the cause of this growth? *

masya89 [10]

Answer:c

Explanation:

4 0

3 years ago

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