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4 years ago

First gets brainliest

Chemistry

1 answer:

mezya [45]4 years ago

5 0

Answer:

reactants : before arrow sign

: CH4 and O2

product : after arrow sign

: CO2 and H2O

the answer is the second option

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Calculate the concentration of OH-in a solution that has a concentration of H+ = 8.1 x 10^−6 M at 25°C. Multiply the answer you

Nat2105 [25]

Answer:

The answer is 12.35

Explanation:

From the question we are given that the concentration of $H^{+}$ is $8.1 * 18^{-6}M$

Generally The rate equation is given as

$K_{w} = [H^{+} ][OH^{-} ]$

and $K_{w}$ the rate constant has a value $1 * 10^{-14}$

Substituting and making [ $OH^{-}$ ] the subject we have

$[OH^{-} ] = \frac{1 * 10^{-14}}{[H^{+}]} = \frac{1 * 10^{-14}}{8.1 *10^{-6}} =1.235 * 10^{-9}$

$[OH ^ {-}] = 1.235 * 10^{-9}M$

Multiply the value by $10^{10}$ as instructed from the question we have

Answer = $1.235 * 10 ^{-9} * 10^{10} = 12.35$

Hence the answer in 2 decimal places is 12.35

7 0

4 years ago

Which term best describes the molecular geometry of ethylene, C₂H₄?

marusya05 [52]

The molecular geometry is trigonal planar. I would choose E

7 0

3 years ago

is the general formula of a certain hydrate. When 256.3 g of the compound is heated to drive off the water, 214.2 g of anhydrous

Marizza181 [45]

Answer:

The general formula of the hydrate is Caa Seb Oc. nH2O. Based on the given information, the weight of the hydrated compound is 256.3 grams, the weight of the anhydrous compound is 214.2 grams.

Therefore, the weight of water evaporated is 256.3 g - 214.2 g = 42.1 grams

The molecular weight of water is 18 gram per mole. So, the number of moles of water will be,

Moles of water = weight of water/molecular weight

= 42.1 grams / 18 = 2.3

The given composition of calcium is 21.90 %. So, the concentration of calcium in anhydrous compound is,

= 214.2 * 0.2190 = 46.91 grams

The given composition of Se is 43.14 %. So, the concentration of selenium in anhydrous compound is,

= 214.2 * 0.4314 = 92.40 grams

The given composition of oxygen is 34.97%, So, the concentration of oxygen in anhydrous compound is,

= 214.2 * 0.3497 = 74.91 grams

The molecular weight of Ca is 40.078, the obtained concentration is 46.91 grams, stoichiometry will be, 46.91/40.078 = 1.17

The molecular weight of Se is 78.96, the obtained concentration is 92.40, stoichiometry will be,

92.40/78.96 = 1.17

The molecular weight of Oxygen is 15.999, the concentration obtained is 74.91, the stoichiometry will be,

74.91/15.999 = 4.68.

Thus, the formula becomes, Ca1.17. Se1.1e O4.68. 2.3H2O, the closest actual component is CaSeO4.2H2O

6 0

4 years ago

What happens to iron oxide during decomposition

elena55 [62]

Answer:

it gets reduced from a +3 oxidation to a 0.

Explanation:

the decomposition of iron oxide to elemenTal can be represented by the following equation: iron oxide (Fe203), the oxidation state of iron is +3 while that of oxygen is -2. therefore, the above reaction is a redox (reduction oxidation reaction)

4 0

3 years ago

29.0 argon combine completely with 4.30 g of Sulfur.

Iteru [2.4K]

Answer:

33.3

Explanation:

if we were to add 29.0 of argon and 4.30 of sulfur and that would come out 33.3.

8 0

3 years ago