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2 years ago

Use the pie section above to answer this question.

Chemistry

1 answer:

Vinvika [58]2 years ago

8 0

Answer:

can you show me a picture of the pie chart?

Explanation:

if i dont see the pie chart then this question isn't possible to answer

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**ASAP**

jek_recluse [69]

the answer is C) . I hope this was helpful!

4 0

3 years ago

Read 2 more answers

How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl

VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

$12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}$

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

$4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}$

Part C:

According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

$13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}$

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

$7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3} }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}$

3 0

3 years ago

A student is determining the density of an unknown metal with a mass of 32.56 g. The student partially fills a graduated cylinde

xenn [34]

Answer:

The metal is Tin (Sn)

Explanation:

The following data were obtained from the question:

Mass of metal = 32.56 g

Volume of water = 14.78 mL

Volume of water + metal = 20.44 mL

Next, we shall determine the volume of the metal. This is illustrated below:

Volume of water = 14.78 mL

Volume of water + metal = 20.44 mL

Volume of metal =..?

Volume of metal = (Volume of water + metal) – (Volume of water )

Volume of metal = 20.44 – 14.78

Volume of metal = 5.66mL

Next, we shall determine the density of the metal.

Density of a substance is defined as the mass of the substance per unit volume of the substance. Mathematically, the density of a substance is expressed as:

Density = Mass / volume

With the above formula, we can obtain the density of the metal as follow:

Mass of metal = 32.56 g

Volume of metal = 5.66mL

Density =.?

Density = Mass /volume

Density = 32.56g/5.66mL

Density of the metal = 5.75g/mL

Comparing the density of the metal with standard density of elements, the metal is Tin (Sn)

4 0

3 years ago

an element "x" has the election configuration 2,8,3 this element is found in ______group and_____period A.3,llla B.llla,3 C.vlla

iogann1982 [59]

This element is found in group 3A, period 3

<h3>Further explanation </h3>

The maximum number of electrons that can be filled in the nth electron shell is 2n²(n=shell)

K shell (n = 1) maximum 2 x 1² = 2 electrons

L shell (n = 2) maximum 2 x 2² = 8 electrons
M shell (n = 3) maximum 2 x 3² = 18 electrons
N shell (n = 4) maximum 2 x 4² = 32 electrons

Electron configuration of element X : 2.8.3 , so :

K shell = 2 ⇒1s²

L shell = 8⇒2s²2p⁶

M shell = 3⇒ 3s²3p¹

Block p: group 13-18 (has a 2p-6p configuration), also called a representative element because it includes metals, non-metals and metalloids

The outer shell 3s²3p¹ : located in group 3A and period 3

group⇒valence electron ⇒3

period⇒the greatest value of the quantum number n⇒3

3 0

3 years ago

muriatic acid is the commercial name for hydrochloric acid that can be purchased from hardware stores as a solution that is 30%(

SSSSS [86.1K]

We have 30% (w/w) HCl this means :
100 g solution contains 30 g HCl
?? g solution contains 7.5 g HCl
By cross multiplication:
Mass of solution = $\frac{100 x 7.5 }{30}$ = 25 g solution

5 0

3 years ago