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Ymorist [56]
3 years ago
8

More points p.s i gave points like 7 times already.... lol hope it helps though

Chemistry
2 answers:
dimulka [17.4K]3 years ago
8 0

Answer:

lol we appreciate you!!

Explanation:

thank you

Anna007 [38]3 years ago
7 0

Appreciation kind of like a good idea

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Balance the following reaction. A coefficient of "1" is understood. Choose the option "blank" for the correct answer if the coef
Charra [1.4K]
C3H8 + 5O2 —-> 3CO2 + 4H2O
8 0
3 years ago
Read 2 more answers
Consider a sample of calcium carbonate in the form of a cube measuring 2.805 in. on each edge.
Naya [18.7K]

Answer:

The answer to your question is: 8.82 x 10 ²⁴ atoms of oxygen          

Explanation:

Data

Cube measuring : 2.805in on each edge

density = 2.7 g/cm3

# of oxygen atoms in the cube = ?

Process

Volume of the cube = (2.805)³ = 22.07 in³

Convert in³ to cm³                  1 in³   -------------------  16.39 cm³

                                       22.07 in³    --------------------    x

                        x = (22.07 x 16.39) / 1 = 361.72 cm³

Density = mass / volume

Mass = density x volume

Mass = (2.7)(361.72) = 976.66 g

Atomic mass CaCO3 = 40 + 12 + (16 x 3) = 100 g

                                100 g of CaCO3 --------------------  48 g of O2

                                976.66 g of CaCO3 ---------------    x

                           x = (976.66 x 48) / 100 = 468.8 g of O2

                              32 g of O2 ----------------------  1 mol

                           468.8 g of O2 --------------------   x

                       x = (468.8 x 1) / 32 = 14.65 mol

                           1 mol -----------------------   6.023 x 10 ²³ atoms

                        14.65 mol -------------------    x

                    x = (14.65 x 6.023 x 10 ²³) / 1 = 8.82 x 10 ²⁴ atoms of oxygen          

3 0
3 years ago
A certain reaction at equilibrium has more moles of gaseous products than of gaseous reactants.(b) Write a statement about the r
mojhsa [17]

To be equal as K_p , K_c needs to be divided by (RT)^{\Delta n}

<h3>Briefly explained</h3>

When the amount of gaseous reactant is greater than the amount of gaseous product, where n_{products} < n_{reactants},

then = K_p < K_c

It's because \Delta n is negative, which places (RT) in the denominator. This is how the equation will now appear.

K_p = K_c(RT)^{-\Delta n} = \frac{ K_c}{(RT)^{\Delta n}}

Here you can observe the value of K_c  is greater than K_p. To be equal as K_p , K_c needs to be divided by (RT)^{\Delta n}

<h3>What is Δngas?</h3>

The number of moles of gas that move from the reactant side to the product side is denoted by the symbol ∆n or delta n in this equation.

Once more, n represents the growth in the number of gaseous molecules the equilibrium equation can represent. When there are exactly the same number of gaseous molecules in the system, n = 0, Kp = Kc, and both equilibrium constants are dimensionless.

<h3>Definition of equilibrium</h3>

When a chemical reaction does not completely transform all reactants into products, equilibrium occurs. Many chemical processes eventually reach a state of balance or dynamic equilibrium where both reactants and products are present.

Learn more about equilibrium

brainly.com/question/11336012

#SPJ4

6 0
2 years ago
A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →
vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

7 0
3 years ago
How many moles of copper are in 6,000,000 atoms of copper?
crimeas [40]

Answer: There will be 9.9632 × 10⁻¹⁸ moles of Copper in 6,000,000 atoms of Copper.

7 0
3 years ago
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