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The amount of space a material occupies is its

SIZIF [17.4K]

Answer:

—MASS is the amount of matter that makes up something. - VOLUME _ is the amount of space that a material takes up.

Explanation:

5 0

3 years ago

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of B

loris [4]

I believe here is the right question, so will just ignore the rest of the junk information from the previous message

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.944 g of H2O(g). Insert subscripts below to appropriately display the empirical formula of BHT

Answer:

$C_{15}H_{24}0$

Explanation:

A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.994 g of H2O(g).

If all the carbon in BHT is present in $CO_2$ and also, all the hydrogen in BHT is present in $H_2O$ , Then we can determine for the corresponding numbers of moles of Carbon(C) and Hydrogen (H) respectively as:

moles of $CO_2$ = $8.990 g*(\frac{1mole}{44.01g})$

= 0.2043 moles

∴ moles of C = 0.2043 moles

moles of $H_2O$ = $2.944 g *(\frac{1mole}{18.01g} )$

= 0.1635 moles

∴ moles of H = 2 × 0.1635 moles

= 0.327 moles

Since number of moles= $\frac{mass}{molarmass}$

number of moles of H = 0.327 moles

molar mass of H = 1.008 g/mol

∴ mass of H in the sample = 0.327 moles × 1.008 g/mol

= 0.329616g

mass of C in the sample can be calculated as = 0.2043 moles × $(\frac{12.01g}{1 mole} )$

= 2.453643 g

mass of C+H in the sample = 2.453643g + 0.329616g

= 2.783259 g

mass of O can be calculated as = 3.001 g - 2.783259 g

= 0.217741 g

∴ moles of O = 0.217741g × $(\frac{1mole}{16.0g})$

= 0.0136 moles

Now, since; we've gotten our data, we can now proceed to calculate for the empirical formula.

C H O

0.2043 0.327 0.0136

Dividing by the least number (0.0136) , we have :

$\frac{0.2043}{0.0136}$ $\frac{0.327}{0.0136}$ $\frac{0.0136}{0.0136}$

15.02 24.04 1

≅

15 24 1

Therefore, the empirical formula would be : $C_{15}H_{24}0$

7 0

3 years ago

Part IV. Limiting Reactants! A Challenge Problem!

Alexxandr [17]

Answer:

a. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)

b. Fe2O3 is the limiting reactant.

c. 6.30 grams Fe

d. 52.6 %

Explanation:

Step 1: Data given

Mass of iron(III) oxide Fe2O3 = 9.00 grams

Mass of aluminium = 4.00 grams

Molar mass Fe2O3 = 159.69 g/mol

Aluminium molar mass = 26.98 g/mol

Step 2: The balanced equation

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)

Step 3; Calculate Moles

Moles = mass / molar mass

Moles Fe2O3 = 9.00 grams / 159.69 g/mol

Moles Fe2O3 = 0.0564 moles

Moles Al = 4.00 grams / 26.98 g/mol

Moles Al = 0.148 moles

Step 4: Calculate limiting reactant

For 1 mol Fe2O3 we need 2 moles Al to produce 2 moles Fe and 1 mol Al2O3

Fe2O3 is the limiting reactant. It will completely be consumed (0.0564 moles). Al is in excess. There will react 0.0564*2 = 0.1128 moles

There will remain 0.148 - 0.1128 = 0.0352 moles Al

Step 5: Calculate moles Fe

For 1 mol Fe2O3 we need 2 moles Al to produce 2 moles Fe and 1 mol Al2O3

For 0.0564 moles Fe2O3 we'll have 2*0.0564 = 0.1128 moles Fe

Step 6: Mass of Fe

Mass Fe = 0.1128 moles * 55.845 g/mol

Mass Fe = 6.30 grams

Step 7: If you carried out this reaction and it actually produced 0.475 mL of molten iron (r = 6.98 g/mL), what is the percent yield of this reaction?

Density = mass / volume

Mass = density * volume

Mass = 6.98 g/mL * 0.475 mL

Mass = 3.3155 grams

Percent yield = (actual mass / theoretical mass) * 100%

Percent yield = (3.3155 /6.30 ) * 100 %

Percent yield = 52.6 %

3 0

3 years ago

What mass of hydrochloric acid (in grams) can 2.7 g of sodium bicarbonate neutralize? (Hint: Begin by writing a balanced equatio

Julli [10]

Answer:

1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate

Explanation:

Step 1: Data given

Mass of sodium bicarbonate = 2.7 grams

Step 2: The balanced equation

HCl + NaHCO3 ⇔ NaCl + H2O + CO2

Step 3: Calculate moles NaHCO3

moles NaHCO3 =2.7 g / 84 g/mol= 0.032 moles

Step 4: Calculate moles HCl

For 1 mol NaHCO3 we need 1 mol HCl

For 0.032 moles NaHCO3 = 0.032 moles HCl

Step 5: Calculate mass HCl

Mass HCl = moles HCl * molar mass HCl

mass HCl = 0.032 * 36.46 g/mol= 1.17 grams

1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate

3 0

3 years ago

Reactions which do not continue to completion are called _____reactions

SIZIF [17.4K]

I think the answer is reversible. Reversible reaction is a kind of reaction which can react in both direction at the same time. And when the both side reactions get the same speed, the reaction is at a equilibrium state.

8 0

3 years ago

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