How do you balance the equation C5H9O +O2 -----> CO2 + H2O

The ability of a substance to combine chemically with another substance.

Marysya12 [62]

Reactivity. It is the ability of matter to combine chemically with other substances.

8 0

2 years ago

What is the density of carbon dioxide gas if 0.196g occupies a volume of 100mL?

-BARSIC- [3]

We know that the equation for density is:

$D=\frac{m}{V}$

where D is the density, m is the mass in grams, and V is the volume.

Given two of the variables, we can then solve for density:

$D=\frac{0.196g}{100mL}= \frac{0.00196g}{mL}$

So therefore, we now know that the density of carbon dioxide gas is 0.00196g/mL.

7 0

3 years ago

Consider the following Reaction.

jeyben [28]

Hey there!:

Molar mass of Mg(OH)2 = 58.33 g/mol

number of moles Mg(OH)2 :

moles of Mg(OH)2 = 30.6 / 58.33 => 0.5246 moles

Molar mass of H3PO4 = 97.99 g/mol

number of moles H3PO4:

moles of Mg(OH)2 = 63.6 / 97.99 => 0.649 moles

Balanced chemical equation is:

3 Mg(OH)2 + 2 H3PO4 ---> Mg3(PO4)2 + 6 H2O

3 mol of Mg(OH)2 reacts with 2 mol of H3PO4 ,for 0.5246 moles of Mg(OH)2, 0.3498 moles of H3PO4 is required , but we have 0.649 moles of H3PO4, so, Mg(OH)2 is limiting reagent !

Now , we will use Mg(OH)2 in further calculation .

Molar mass of Mg3(PO4)2 = 262.87 g/mol

According to balanced equation :

mol of Mg3(PO4)2 formed = (1/3)* moles of Mg(OH)2

= (1/3)*0.5246

= 0.1749 moles of Mg3(PO4)2

use :

mass of Mg3(PO4)2 = number of mol * molar mass

= 0.1749 * 262.87

= 46 g of Mg3(PO4)2

Therefore:

% yield = actual mass * 100 / theoretical mass

% = 34.7 * 100 / 46

% = 3470 / 46

= 75.5%

Hope that helps!

3 0

3 years ago

During autumn, days grow colder. What happens to the nights?

Rus_ich [418]

Answer:

I thinks its they grow longer explaination: I have a massive brain

7 0

2 years ago

Read 2 more answers

A lead ball is added to a graduated cylinder containing 50.6 ml of water, causing the level of the water to increase to 93.0 mL.

Kamila [148]

42.4 ml is the volume in milliliters of the lead ball if a lead ball is added to a graduated cylinder containing 50.6 ml of water.

<h3>What is a graduated cylinder?</h3>

A tall narrow container with a volume scale is used especially for measuring liquids.

The graduated cylinder contains water

mL is a volume unit.

Water volume = 50.6 ml

The lead ball caused an increase in volume from 50.6 ml to 93.0 mL.

The new volume is the lead ball volume plus the original water volume :

Final volume = Vlead ball+ Water original volume

$93.0 mL= V_(lead ball) +50.6 ml$

$V_(lead ball) = 93.0 mL - 50.6 ml$

$V_(lead ball) = 42.4 ml$

Hence, 42.4 ml is the volume in milliliters of the lead ball.

Learn more about the graduated cylinder here:

brainly.com/question/13386106

#SPJ1

4 0

2 years ago