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3 years ago

Use the drop-down menus to determine which percentage is equal to each fraction.

Chemistry

2 answers:

kondor19780726 [428]3 years ago

8 0

1/4: 25%
3/4: 75 %
4/4: 100%
2/4: 50%

Hope this helped :)

Snowcat [4.5K]3 years ago

7 0

Answer: these are the answers i got.

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Name the type of subunits that form starch

Natasha2012 [34]

Starch is a polysaccharides which is made up of a long chain of glucose. The sub units that for starch are GLUCOSE. The long chains of glucose in starch are connected by alpha 1,4 linkages. The simplest starch in existence is amylose.

5 0

3 years ago

What volume of nitrogen (n2) would be completely consumed in the reaction with 30.80 g of

Shtirlitz [24]

The answer is 285.33g nitrogen would be completely consumed in the reaction with 30.80 g of hydrogen gas.

A mole is defined as 6.02214076 × 10²³ atoms, molecules, ions, or other chemical units.

Write a balanced equation.

Calculate the moles of H₂ in 30.8 g.

Calculate the moles of N₂ required to react with H₂.

Calculate the mass of N₂.

Calculate the initial mass of N₂.

Start with a balanced equation.

N₂ + 3H₂ --> 2NH₃

Calculate the moles of H₂ in 30.8 g.

n = m/M; where n = moles, m = mass, and M = molar mass.

M(H₂) = 1.008 g/mol

n(H₂) = (30.8 g)/(1.008 g/mol) = 30.56 mol H₂

Calculate the moles of N₂ required to react with 30.56 mol H₂ , using the mole ratio between H₂ and N₂ in the balanced equation.

30.56 mol H₂ × 1 mol N₂/3 mol H₂ = 10.18 mol N₂

Calculate the mass of N₂ in 10.18 mol.

m = n × M

M(N₂) = 2 × 14.007 g/mol N = 28.014 g/mol N₂

m(N₂) = 10.18 mol × 28.014 g/mol = 285.33g N₂

Therefore 285.33g nitrogen would be completely consumed in the reaction with 30.80 g of hydrogen gas.

To know more about mole

brainly.com/question/26416088

#SPJ1

5 0

2 years ago

A standard gold bar has a mass of 12.4 kg. When the gold bar is placed into a container of water the volume changes by 642 mL. W

AlekseyPX

Answer:

19.3 g/mL

Explanation:

The following data were obtained from the question:

Mass (m) = 12.4 kg

Volume (V) = 642 mL.

Density (D) =.?

Next, we shall convert 12.4 kg to grams. This can be obtained as follow:

1 kg = 1000 g

Therefore,

12.4 kg = 12.4 × 1000

12.4 kg = 12400 g

Therefore, 12.4 kg is equivalent to 12400 g.

Finally, we shall determine the density of the gold as follow:

Density is simply defined as the mass of the substance per unit volume of the substance. It can be represented mathematically as:

Density (D) = mass (m) / volume (V)

D = m/V

With the above formula, the density of gold can easily be obtained as follow:

Mass (m) = 12400 g

Volume (V) = 642 mL.

Density (D) =.?

D = m/V

D = 12400/642

D = 19.3 g/mL

Therefore, the density of hold is 19.3 g/mL

6 0

3 years ago

When the temperature of a substance decreases to the substance's freezing point, the substance will begin to change from a _____

igor_vitrenko [27]

It changes from a liquid to a solid.

4 0

4 years ago

The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

enot [183]

Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}$

7 0

3 years ago