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2 years ago

A beaker is best used for measuring

Chemistry

2 answers:

sergey [27]2 years ago

3 0

liquids, it has the measurements in ml.

fenix001 [56]2 years ago

3 0

liquids is a beaker is used for mesuring liquids

-QUEENSHADYGUCCI-

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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

Use the balanced equation to solve the problem.

harkovskaia [24]

Answer:

4.20 moles NF₃

Explanation:

To convert between moles of N₂ and NF₃, you need to use the mole-to-mole ratio from the balanced equation. This ratio consists of the coefficients of both molecules from the balanced equation. The molecule you are converting from (N₂) should be in the denominator of the ratio because this allows for the cancellation of units. The final answer should have 3 sig figs because the given value (2.10 moles) has 3 sig figs.

1 N₂ + 3 F₂ ---> 2 NF₃

2.10 moles N₂ 2 moles NF₃
--------------------- x --------------------- = 4.20 moles NF₃
1 mole N₂

7 0

1 year ago

(a) what is the approximate ductility (%el) of a brass that has a yield strength of 230 mpa (33360 psi)?

ss7ja [257]

Based on the calculations, the approximate ductility (%el) of this brass is equal to 2.3%.

<u>Given the following data:</u>

Yield strength = 230 mpa (33360 psi).

<h3>What is ductility?</h3>

Ductility can be defined as an important property of a material which determines its ability to become elongated due to the application of stress.

Mathematically, the ductility of a material can be expressed as percentage elongation in length:

$\% el = \frac{\Delta L}{L_i} =\frac{L_f - L_i}{L_i}$