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suter [353]
3 years ago
10

The equation h3po4 + 3koh →k3po4 + 3h2o is an example of which type of reaction?

Chemistry
2 answers:
Doss [256]3 years ago
5 0

double replacement i took the test

faust18 [17]3 years ago
3 0
<span>b. double-replacement not sure 
</span>
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4 years ago
A sample of an ideal gas at 1.00 atm and a volume of 1.72 l was placed in a weighted balloon and dropped into the ocean. as the
kolbaska11 [484]
If we analyze the situation analytically, there are situations or states. Then, we are also given with 2 values of pressure and 1 value of volume. Lastly, temperature was set as constant. Thus, this means we use the Boyle's Law.

P₁V₁ = P₂V₂
Let's find V₂.
(1 atm)(1.72 L) = (35 atm)(V₂)
Solving for V₂,
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7 0
3 years ago
QUESTION 4 Which structure is found in all eukaryotic cells? large central vacuole Golgi apparatus flagella cilia
jok3333 [9.3K]
I believe it's the Golgi Apparatus
5 0
4 years ago
Read 2 more answers
What mass of H₂ is needed to react with 8.75 g of O₂ according to the following equation: O2(g) + H2(g) → H₂O(g)?
FromTheMoon [43]

Explanation:

For reacting with 8.75 grams of oxygen, 1.08 grams of hydrogen is required.

The given balanced equation has been:

\rm O_2\;+\;2\;H_2\;\rightarrow\;H_2OO2+2H2→H2O

From the equation, 1 mole of oxygen reacts with 2 mole of hydrogen to give 1 mole of water.

The mass of oxygen has been: 8.75 g,

Moles = \rm \dfrac{weight}{molecular\;weight}molecularweightweight

Moles of oxygen = \rm \dfrac{8.75}{32}328.75

Moles of oxygen = 0.27 mol

Since,

1 mole Oxygen = 2 mole hydrogen

0.21 mol oxygen = 0.54 mol hydrogen

Mass of hydrogen = moles \times× molecular weight

Mass of hydrogen = 0.54 \times× 2

Mass of hydrogen = 1.08 grams.

Thus, for reacting with 8.75 grams of oxygen, 1.08 grams of hydrogen is required.

6 0
2 years ago
The average bond energy (enthalpy) for a C=C double bond is 614 kJ/mol and that of a C−C single bond is 348 kJ/mol. Estimate the
AysviL [449]

Answer:

4.42x10⁻¹⁹ J/molecule

Explanation:

At a double bond, there's sigma and a pi bond, and at a single bond, there's only a sigma bond. Thus, if the energy to break both sigma and pi is 614 kJ/mol, and the energy to break only the sigma bond is 348 kJ/mol, the energy to break only the pi bond is:

E = 614 - 348 = 266 kJ/mol

Knowing that 1 kJ = 1000 J, E = 266,000 J/mol

By Avogadro's number, 1 mol = 6.02x10²³ molecules, thus:

E = 266,000 J/mol * 1mol/6.02x10²³ molecules

E = 4.42x10⁻¹⁹ J/molecule

7 0
3 years ago
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