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3 years ago

How is compound different from a mixture?

Chemistry

2 answers:

Afina-wow [57]3 years ago

8 0

I think it should be <span>D. it cannot be separated by a physical change</span>

kenny6666 [7]3 years ago

3 0

A compound is chemically bonded together atoms and a mixture is two substances combined, i think its D

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What is the volume occupied by 2.50 moles of oxygen gas at STP

oee [108]

1 mole of any gas under STP has volume 22.4 L
So 2.50 moles of any gas ( including oxygen)
2.50 mol *(22.4L/1 mol)=56.0 L

8 0

2 years ago

What isapplied in case where the base is not a OH donator

blagie [28]

It will received an H+. I’m not sure if that answers your question correctly

8 0

2 years ago

I am stuck on this lab and have absolutely no idea how to do it.

expeople1 [14]

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4 0

3 years ago

Read 2 more answers

If the CaCO3 weighed 983 g and the CaO weighed 551 g, how many grams of CO2 were formed in the reaction?

stira [4]

Answer:

The answer to your question is 432 g of CO₂

Explanation:

Data

CaCO₃ = 983 g

CaO = 551 g

CO₂ = ?

Balanced reaction

CaCO₃ (s) ⇒ CaO (s) + CO₂ (g)

This reaction is balanced, to solve this problem just remember the Lavoisier Law of conservation of mass that states that the mass of the reactants is equal to the mass of the products.

Mass of reactants = Mass of products

Mass of CaCO₃ = Mass of CaO + Mass of CO₂

Solve for CO₂

Mass of CO₂ = Mass of CaCO₃ - Mass of CaO

Mass of CO₂ = 983 g - 551 g

Simplification

Mass of CO₂ = 432 g

5 0

2 years ago

If is found that 24.68 mL of .1165 M NaOH is needed to titrate .2931 g of an unknown acid to the phenolphthalein end point. Calc

Lemur [1.5K]

Answer: The equivalent mass of the acid is 83.16 grams

Explanation:

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

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

Molarity of $NaOH$ solution = 0.1165 M

Volume of $NaOH$ solution = 24.68 mL = 0.02468 L

Putting values in equation 1, we get:

$\text{Moles of} NaOH={0.1165M}\times{0.02468L}=2.875\times 10^{-3}moles=2.875\times 10^{-3}geq$

( as acidity of NaOH is 1)

For end point: gram equivalents of acid = gram equivalents of base = $2.875\times 10^{-3}$

Mass of acid= $gram equivalents\times {\text {Equivalent mass}}$

$0.2391=2.875\times 10^{-3}\times {\text {Equivalent mass}}$

${\text {Equivalent mass}}=83.16g$

Thus equivalent mass of the acid is 83.16 grams

5 0

3 years ago