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3 years ago

Pls help!! will give brainly

Chemistry

2 answers:

lutik1710 [3]3 years ago

6 0

Definitely true! Science is based on continuing research and evidence and therefore constantly changing.

____ [38]3 years ago

4 0

Answer:

True

Explanation:

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For the problem: Mg (s) + O₂ (g) ➞ MgO (g) if 4.3 g of magnesium reacts with 83 g of oxygen (O₂), which is the limiting reagent?

Anit [1.1K]

Explanation:

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3 0

3 years ago

Based on the balanced equation 2C2H2 + 5O2 → 4CO2 + 2H2O calculate the number of excess reagent units remaining when 52 C2H2 mol

erica [24]

Answer:

20 molecules of oxygen gas remains after the reaction.

Explanation:

$2C_2H_2 + 5O_2\rightarrow 4CO_2 + 2H_2O$

Molecules of ethyne = 52

Molecules of oxygen gas = 150

According to reaction, 2 molecules of ethyne reacts with 5 molecules of oxygen gas.

Then 52 molecules of ethyne will react with:

$\frac{5}{2}\times 52=130 molecules$ of oxygen gas.

As we can see that we have 150 molecules of oxygen gas, but 52 molecules of ethyne will react with 130 molecules of oxygen gas. So, this means that ethyne is a limiting reagent and oxygen gas is an excessive reagent.

Remaining molecules of recessive reagent = 150 - 130 = 20

20 molecules of oxygen gas remains after the reaction.

5 0

3 years ago

What is 450 g of flour a measure of?

Marat540 [252]

The right answer is D - mass.

3 0

3 years ago

Read 2 more answers

Following the instructions in your lab manual, you have titrated a 25.00 mL sample of 0.0100 M KIO3 with a solution of Na2S2O3 o

vivado [14]

Answer:

For 1: The amount of potassium iodate that were titrated is $2.5\times 10^{-4}$ moles

For 2: The amount of sodium thiosulfate required is $1.25\times 10^{-4}$ moles

Explanation:

For 1:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$

Molarity of $KIO_3$ solution = 0.0100 M

Volume of solution = 25 mL

Putting values in above equation, we get:

$0.0100M=\frac{\text{Moles of }KIO_3\times 1000}{25}\\\\\text{Moles of }KIO_3=\frac{0.0100\times 25}{1000}=0.00025mol$

Hence, the amount of potassium iodate that were titrated is $2.5\times 10^{-4}$ moles

For 2:

The chemical equation for the reaction of potassium iodate and sodium thiosulfate follows:

$2KIO_3+Na_2S_2O_3\rightarrow K_2S_2O_3+2NaIO_3$

By Stoichiometry of the reaction:

2 moles of potassium iodate reacts with 1 mole of sodium thiosulfate

So, 0.00025 moles of potassium iodate will react with = $\frac{1}{2}\times 0.00025=0.000125mol$ of sodium thiosulfate

Hence, the amount of sodium thiosulfate required is $1.25\times 10^{-4}$ moles

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3 years ago

What kind of scientist would study the the first of a car running

iVinArrow [24]

Answer:

A physicist

Explanation:

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7 0

3 years ago