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7nadin3 [17]
3 years ago
9

The statement "matter can be neither created nor destroyed by chemical means, but it can be changed from one form to another" is

the chemical law of
CONSERVATION OF MATTER
Chemistry
1 answer:
Assoli18 [71]3 years ago
4 0
I think false because matter cant be changed into something it isnt.<span />
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Phantasy [73]
Your answer would be 58.12g/mol ;)
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A chemist adds 485 mL of a 0.0025 mol/L calcium sulfate solution to a reaction flask. Calculate the mass in grams of calcium sul
viva [34]

Answer : The mass in grams of calcium sulfate is 0.16 grams.

Explanation :

Molarity : It is defined as the number of moles of solute present in one litre of solution.

Formula used :

Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution}}

Solute is, CaSO_4

Given:

Molarity of CaSO_4 = 0.0025 mol/L

Molar mass of CaSO_4 = 136 g/mole

Volume of solution = 485 mL

Now put all the given values in the above formula, we get:

0.0025=\frac{\text{Mass of }CaSO_4\times 1000}{136\times 485}

\text{Mass of }CaSO_4=0.16g

Thus, the mass in grams of calcium sulfate is 0.16 grams.

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3 years ago
What is the use of oxyacetylene flame
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Answer:

An oxyacetylene torch can also be used for welding. When welding with an oxyacetylene torch, the flame is used to produce molten metal along the edge of two work pieces.

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7 0
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3 years ago
Calculate ΔG o for the following reaction at 25°C: 3Mg(s) + 2Al3+(aq) ⇌ 3Mg2+(aq) + 2Al(s) Enter your answer in scientific notat
larisa [96]

Answer:

-3.7771 × 10² kJ/mol

Explanation:

Let's consider the following equation.

3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)

We can calculate the standard Gibbs free energy (ΔG°) using the following expression.

ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)

where,

n: moles

ΔG°f(): standard Gibbs free energy of formation

p: products

r: reactants

ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))

ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)

ΔG° = -377.71 kJ = -3.7771 × 10² kJ

This is the standard Gibbs free energy per mole of reaction.

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