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bezimeni [28]
3 years ago
15

Im literally gonna cry . what is this

Chemistry
2 answers:
Vlad1618 [11]3 years ago
6 0
Ok I will!!!!!!!!!!!
Dvinal [7]3 years ago
5 0
What is the picture lol , it looks like it was took on a calculator....
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Calculate AH for the reaction:<br> 2N2 (g) + 6H20 (g) → 302 (g) + 4NH3(g)
Ratling [72]

Answer:

Explanation:

4NH₃ (g) + 3O₂ (g) ⇒ 2N₂ (g) + 6H₂ O(1)

ΔH_r={\sum}{ \Delta }H_f(\text{product})-{\sum}{ \Delta }H_f(\text{reactant})

ΔH r =(2ΔH f(N 2 )+6ΔH f (H 2 O(l)))−(4ΔH f​ (NH 3 (g))+3ΔH f (O 2 (g)))

ΔH rex =[2×0+6×(−286)]−[4×(−46)+3×0]=−1716+186

ΔH rex =−1532kJ/mol

Thermodynamics is a branch of physical chemistry that studies heat and its effects and interactions. Governed by the four main laws, thermodynamics plays a huge role in physics and chemistry, and is also responsible for the law of conservation of energy, a fundamental rule in science.

7 0
2 years ago
What is an item or substance that would react with hydrochloric
vovikov84 [41]
Magnesium(?)
<span>2 HCl + Mg ? MgCl2 + H2</span>
4 0
3 years ago
Read 2 more answers
What temperature in degrees c is the freezing point of water?
guajiro [1.7K]
The freezing point of water is 0° C. The Celsius (centigrade) scale is based off of water, with the freezing point at 0° and the boiling point at 100°. (Google will probably tell you this in 10 seconds, then you wouldn't have had to wait 7 hours!)
5 0
3 years ago
For the following reaction, 38.3 grams of sulfuric acid are allowed to react with 33.5 grams of calcium hydroxide sulfuric acid(
Likurg_2 [28]

Answer:

What is the maximum amount of calcium sulfate that can be formed? 53.1 grams CaSO4

What is the FORMULA for the limiting reagent? H2SO4

What amount of the excess reagent remains after the reaction is complete? 4.59 grams of Ca(OH)2

Explanation:

Step 1: Data given

Mass of sulfuric acid = 38.3 grams

Molar mass of H2SO4 = 98.08 g/mol

Mass of calcium hydroxide = 33.5 grams

Molar mass of Ca(OH)2 = 74.09 g/mol

Step 2: The balanced equation

H2SO4 + Ca(OH)2 → CaSO4 + 2H2O

Step 3: Calculate moles of H2SO4

moles H2SO4 = mass H2SO4 / molar mass H2SO4

moles H2SO4 = 38.3 grams / 98.08 g/mol

moles H2SO4 = 0.390 moles

Step 4: Calculate moles of Ca(OH)2

moles Ca(OH)2 = 33.5 grams / 74.09 g/mol

moles Ca(OH)2 =0.452 moles

Step 5: Calculate limiting reactant

For 1 mol H2SO4, we need 1 mol of Ca(OH)2 to produce, 1 mol of CaSO4 and 2 mol of H2O

H2SO4 is the limiting reactant. It will completely be consumed (0.390 moles).

Ca(OH)2 is in excess. There will be consumed 0.390 moles

There will remain 0.452 - 0.390 = 0.062 moles

This is 0.062 * 74.09 g/mol = 4.59 grams

Step 6: Calculate moles of calcium sulfate

For 1 mol H2SO4, we need 1 mol of Ca(OH)2 to produce, 1 mol of CaSO4 and 2 mol of H2O

For 0.390 moles of H2SO4, there will be produced 0.390 moles of CaSO4

Step 7: Calculate mass of CaSO4

Mass CaSO4 = moles CaSO4 * molar mass CaSO4

Mass CaSO4 = 0.390 moles * 136.14 g/mol

Mass of CaSO4 = 53.1 grams

7 0
3 years ago
A 10.5 mL sample of vinegar, containing acetic acid, was titrated using 0.460 M NaOH solution. The titration required 19.13 mL o
laila [671]

Explanation:

Step 1:

A good first step for a problem like this is to write down the chemical formula and balance it.

It appears here that we have 10.5 mL of vinegar, which IS acetic acid, and 19.13 mL of 0.460 M NaOH. That will give us the following balanced chemical equation:

CH3COOH + NaOH ------> NaCH3COO + H2O

All of the constituents come out to a value of 1, conveniently.

Step 2:

Since all of our stoichiometric coefficients are one, we can use a shortcut to answer this equation. I don't know if it has a name, but I just call it the titration formula. It goes something like this:

M1 * V1 = M2 * V2

M stands for Molarity and V stands for volume. 1 and 2 being the before the reaction and after the reaction.

So, our M1 for this is going to be what the question says was used for this titration. That's 0.460M NaOH.

Our V1 is going to be the initial volume of the sample, which was 10.5 mL

Our V2 is going to be 19.13, which is the volume when we're finished.

It's clear that we don't know M2, so let's find it.

Keep in mind that it's easier to convert to liters pretty much always, so I've done that by dividing the mL values each by 1000.

Using some algebra, we can see that we now have:

0.460 M * 0.0105 L = x M * 0.01913 L

Which goes to:

\frac{0.00483mol}{0.01913L} = 0.252 M

<h3>So our M2, the molar concentration of acetic acid in this vinegar, is equal to 0.252 M. </h3>
3 0
3 years ago
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