All categories

3 years ago

Pepsi has a density of 1.06 g/mL. It has a mass of 370 g. What is the volume of the can of Pepsi?

Chemistry

1 answer:

Ulleksa [173]3 years ago

5 0

Answer:

<h3>The answer is 320.75 mL</h3>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density} \\$

From the question we have

$volume = \frac{340}{1.06} \\ = 320.7547169...$

We have the final answer as

Hope this helps you

You might be interested in

If you burn 29.4 g of hydrogen and produce 263 g of water, how much oxygen reacted?

vovikov84 [41]

The reaction between hydrogen and oxygen to form water is given as:

$H_2 + O_2 \rightarrow H_2O$

The balanced reaction is:

$2H_2 + O_2 \rightarrow 2H_2O$

According to the balanced reaction,

4 g of hydrogen ( $4\times 1$ ) reacts with 32 g of oxygen ( $2\times 16$ ).

So, oxygen reacted with 29.4 g of hydrogen is:

$\frac{29.4\times 32}{4} = 235.2 g$

Hence, the mass of oxygen that is reacted with 29.4 g of hydrogen is 235.2 g.

7 0

3 years ago

Calcium cyclamate, Ca(C₆H₁₁NHSO₃)₂, is an artificial sweetener used in many countries around the world but is banned in the Unit

xeze [42]

Answer:

2HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃;

Ba(C₆H₁₁NHSO₃)₂ + H₂SO₄ → BaSO₄ + 2HC₆H₁₁NHSO₃;

2HC₆H₁₁NHSO₃ + Ca(OH)₂ → Ca(C₆H₁₁NHSO₃)₂ + 2H₂O

Explanation:

First, let's see the reactants for the first reaction and how they dissociate:

HC₆H₁₁NHSO₃ → H⁺ + C₆H₁₁NHSO₃⁻

BaCO₃ → Ba²⁺ + CO₃²⁻ (Barium is from group 2, so its cation has charge +2)

So, to form the products, the cation of one will join the anion of others. The amount of the cation will be the charge of the anion, and the amount of the anion will be the charge of the cation:

H⁺ + CO₃²⁻ → H₂CO₃

Ba²⁺ + C₆H₁₁NHSO₃⁻ → Ba(C₆H₁₁NHSO₃)₂

The reaction then is:

HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃

The number of elements must be the same on both sides, so the balanced equation is

2HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃

The treatment with H₂SO₄ will produce:

H₂SO₄ → 2H⁺ + SO₄⁻²

Ba(C₆H₁₁NHSO₃)₂ → Ba²⁺ + C₆H₁₁NHSO₃⁻

The balanced reaction will be then:

Ba(C₆H₁₁NHSO₃)₂ + H₂SO₄ → BaSO₄ + 2HC₆H₁₁NHSO₃

In the last step, HC₆H₁₁NHSO₃ will react with Ca(OH)₂

HC₆H₁₁NHSO₃ → H⁺ + C₆H₁₁NHSO₃⁻

Ca(OH)₂ → Ca²⁺ + 2OH⁻

The balance reaction will be:

2HC₆H₁₁NHSO₃ + Ca(OH)₂ → Ca(C₆H₁₁NHSO₃)₂ + 2H₂O

3 0

3 years ago

2H2 + 02>2 H20<br> What is the reaction

Murrr4er [49]

Answer: The actual reaction to make water is a bit more complicated: 2H2 + O2 = 2H2O + Energy. In English, the equation says: To produce two molecules of water (H2O), two molecules of diatomic hydrogen (H2) must be combined with one molecule of diatomic oxygen (O2). Energy will be released in the process.

Explanation:

8 0

3 years ago

Mass of a mole is known as

gayaneshka [121]

Answer:

Molar mass is the mass of a given substance divided by the amount of that substance, measured in g/mol. For example, the atomic mass of titanium is 47.88 amu or 47.88 g/mol. In 47.88 grams of titanium, there is one mole, or 6.022 x 1023 titanium atoms.

Explanation:

Could I have branliest , heart, and 5 stars

Thanks!

7 0

2 years ago

Write a balanced equation and determine Eº for each of the following cells: a) Cr Cr3+||Ni2+|Ni b) (CF|C1,||MnO | Mn?)

Alexxx [7]

Answer:

Explanation:

Step 1: Write both half reactions

Cr / Cr3+ : oxidation Cr(s) → Cr3+ + 3e-

Ni2+ / Ni : reduction Ni2+ +2e- → Ni

Step 2: Balance reactions and look up the standard potential for the half-reactions

2(Cr → Cr3+ + 3e-) E° ox = 0.74 V

3(Ni2+ +2e- → Ni) E° red = -0.25 V

2Cr + 3Ni2+ +6e- → 2Cr3+ +6e- + 3Ni

E°cell = E° red + E° ox = -0.25 + 0.74 = 0.49

E = E ° − 0.0257 V /n * ln Q = E ° − 0.0257 V /n *l n [ C r 3 + ]/ [ N i 2 + ]

With E° = 0.49 V

Step 1: Write both half reactions

MnO4-/ Mn2+ (redution) MnO4- +8H+ +5e- ⇔ Mn2+ +4H2O

Cf ⇔ Cf2+ +2e- (oxidation) Cf ⇔ Cf2+ +2e-

Step 2: Balance reactions and look up the standard potential for the half-reactions

MnO4- +8H+ +10-e- ⇔ Mn2+ +4H2O E° = 1.51 V

5Cf ⇔ 5 Cf2+ +10e- E° =2.12 V

2 MnO4- + 16H+ + 5Cf ⇔ 2Mn2+ + 8H2O + 5Cf2+

E°cell = E° red + E° ox = 1.51 + 2.12 = 3.63 V

E = E ° − 0.0257 V /n * ln Q = E ° − 0.0257 V /n *l n [ C f2 + ]/ [ Mno4- ]

With E° =3.63 V

7 0

2 years ago