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3 years ago

What is the mole ratio of butane to carbon dioxide?

1 answer:

5 0

Butane is 2, and carbon dioxide is 8. which makes the mole ratio of butane to carbon dioxide 2:8, which simplifies to 1:4.

You might be interested in

How many grams of sodium chloride are present in a 0.75 m solution with a volume of 500.0 milliliters?

nlexa [21]

22grams my boi because tgfghjkhgffdjgg i found it

7 0

3 years ago

If the pressure on a gas is increased, then its volume will...?

oksian1 [2.3K]

Decrease. I got you

4 0

3 years ago

How would you prepare 500 ml of the following solutions : Sodium succinate buffer (0.1 mol/dm3 pH 5.64)

Nana76 [90]

Answer:

8.10g of sodium succinate must be added and 247mL of 0.1M HCl adding enough water until make 500mL

Explanation:

Succinic acid has a pKa₂ of 5.63

To solve this question we must find the amount of sodium succinate and 0.1M HCl that we have to add using H-H equation:

pH = pKa + log [A-] / [HA]

5.64 = 5.63 + log [Na₂Succ.] / [HSucc⁺]

0.01 = log [Na₂Succ.] / [HSucc⁺]

1.0233 = [Na₂Succ.] / [HSucc⁺] (1)

As:

0.1M = [Na₂Succ.] + [HSucc⁺] (2)

Replacing (2) in (1):

1.0233 = 0.1M - [HSucc⁺] / [HSucc⁺]

1.0233[HSucc⁺] = 0.1M - [HSucc⁺]

2.033[HSucc⁺] = 0.1M

[HSucc⁺] = 0.0494M

[Na₂Succ] = 0.0506M

Both [Na₂Succ⁺] and [HSucc⁺] ions comes from the same sodium succinate we have to find the moles of sodium succinate in 500mL of 0.1M. Then, based on the reaction:

Na₂Succ + HCl → HSucc⁺ + Cl⁻

The moles of HCl added = Moles HSucc⁺ we need:

Moles Na₂Succ:

0.500L * (0.1mol/L) = 0.0500 moles

Mass -Molar mass sodium succinate: 162.05g/mol-:

0.0500mol * (162.05g/mol) = 8.10g of sodium succinate must be added

Moles HCl:

0.0494M * 0.500L = 0.0247 moles HCl * (1L / 0.1mol) = 0.247L =

And 247mL of 0.1M HCl adding enough water until make 500mL

7 0

3 years ago

A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num

-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

$V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}$

$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}$

B) Radius

$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}$

$V = \frac{ 4}{3 }\pi r^{3}$

$r^{3} = \frac{3V }{4 \pi }\$

$r= \sqrt [3]{ \frac{3V }{4 \pi } }$

$r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}$

3 0

3 years ago

What is meant by the term chemical reaction

pantera1 [17]

Answer:

<h3> $\huge{ \underline{ \underline { \bold{ \sf{ \orange{See \: below}}}}}}$ </h3>

Explanation:

▪️ $\underline{ \bold{ \sf{ \blue{Chemical \: reaction}}}}$

⇒The composition , decomposition or displacement of molecules of matter during chemical change is called chemical reaction.

▪️ $\underline{ \bold{ \sf{ \purple{ \: Further \: more \: explanation}}}}$

Various conditions bring about these changes. The chemical reactions are represented by chemicalequation. The compounds or elements that take part in chemical reaction are called reactant. They are written at the left side of an arrow that represent a change while the compound or elements that formed after the chemical change are called product. They are written at the right side of the arrow.

▪️ $\underline{ \bold{ \sf{ \purple{For \: example}}}}$

When nitrogen reacts with hydrogen to form ammonia :

Nitrogen + Hydrogen ⇒ Ammonia

N₂ + 3H₂ ⇒ 2NH₃

Presentation of a chemical reaction in the form of equation is called chemical equation . Chemical equation may be word equations or formula equations.

Hope I helped!

Best regards!!

3 0

3 years ago