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MArishka [77]
2 years ago
6

QUICK PLEASE HELP ONLY 2 MINUTES A group of engineers has a problem. Solving this problem requires the engineers to generate man

y different ideas. What creative process would best accomplish this goal?
Chemistry
1 answer:
Mariana [72]2 years ago
8 0

I would say a Think Tank would be the best creative process to help engineers generate ideas. Think Tanks are groups of people that help generate ideas to solve problems and grow companies.

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Dinitrogen pentoxide gas is formed by the reaction of nitrogen trioxide gas and nitrogen dioxide gas balanced equation
leonid [27]

Answer:

Balanced chemical  equation:

NO₂ + NO₃ → N₂O₅

Explanation:

Chemical equation:

NO₂ + NO₃ → N₂O₅

Balanced chemical  equation:

NO₂ + NO₃ → N₂O₅

Nitrogen trioxide gas combine with nitrogen dioxide gas and form nitrogen pentoxide.

This is the simple synthesis reaction in which two substance combine to form a new substance.

Synthesis reaction:

It is the reaction in which two or more simple substance react to give one or more complex product.

General chemical equation:

A + B  → AB

A and B are reactants that combine to form AB product.

7 0
3 years ago
Calculate the mass, in grams, of Ag2CrO4 that will precipitate when 50.0mL of 0.20M AgNO3 solution is mixed with 40.0mL of 0.10M
Darina [25.2K]

Answer:

1.327 g Ag₂CrO₄

Explanation:

The reaction that takes place is:

  • 2AgNO₃(aq) + K₂CrO₄(aq)  → Ag₂CrO₄(s) + 2KNO₃(aq)

First we need to <em>identify the limiting reactant</em>:

We have:

  • 0.20 M * 50.0 mL = 10 mmol of AgNO₃
  • 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄

If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.

Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:

  • 4 mmol K₂CrO₄ * \frac{1mmolAg_2CrO_4}{1mmolK_2CrO_4} *\frac{331.73mg}{1mmolAg_2CrO_4} = 1326.92 mg Ag₂CrO₄
  • 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄
7 0
2 years ago
A 215-g sample of copper metal at some temperature is added to 26.6 g of water. The initial water temperature is 22.22 oC, and t
andrezito [222]

The initial temperature of the copper metal was 27.38 degrees.

Explanation:

Data given:

mass of the copper metal sample = 215 gram

mass of water = 26.6 grams

Initial temperature of water = 22.22 Degrees

Final temperature of water = 24.44 degrees

Specific heat capacity of water = 0.385 J/g°C

initial temperature of copper material , Ti=?

specific heat capacity of water = 4.186 joule/gram °C

from the principle of:

heat lost = heat gained

heat gained by water is given by:

q water = mcΔT

Putting the values in the equation:

qwater = 26.6 x 4.186 x (2.22)

qwater = 247.19 J

qcopper = 215 x 0.385 x (Ti-24.4)

              = 82.77Ti - 2019.71

Now heat lost by metal = heat gained by water

82.77Ti - 2019.71 = 247.19

Ti = 27.38 degrees

8 0
3 years ago
Calculate the ph of a solution containing 0.0451 m potassium hydrogen tartrate and 0.028 m dipotassium tartrate. The ka values f
Marina86 [1]

Given buffer:

potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )

[KHC4H4O6] = 0.0451 M

[K2C4H4O6] = 0.028 M

Ka1 = 9.2 *10^-4

Ka2 = 4.31*10^-5

Based on Henderson-Hasselbalch equation;

pH = pKa + log [conjugate base]/[acid]

where pka = -logKa

In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2

pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]

     = -log (4.31*10^-5) + log [0.0451]/[0.028]

pH = 4.15



4 0
2 years ago
calculate the charge on plates if the plate area is 100 cm2 , the gap between plates is 10 mm and the scale is reading 1.0 gram.
Law Incorporation [45]

Answer: The charge on the plates are 88.4 picafarad

Explanation:The equation used in measuring charge in a plate is given as:

C=Q/V =E A/D

Where E= dielectric content

A= Area of plates

d= distance between plates

Using dielectric constant for Air=8.84×10-12F/m

A=100cm2=0.01m2

d=10mm=0.001m

C= 8.84×10-12×0.01/0.001

C= 88.4 picafarad

8 0
3 years ago
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