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iren [92.7K]
3 years ago
6

How do the ongoing processes of scientific investigation and technological design interact with each other?

Chemistry
2 answers:
Maru [420]3 years ago
8 0

Answer:

The correct answer is option a, that is, scientific investigation discovers new information that can be used in technological design, and technological design provides new products that can further scientific investigation.

Explanation:

Both science and technology possesses distinct objectives, ways of knowing and observing the world, and therefore, their association is usually complex and tense. Though, they work to extend and inform each other in both the unexpected and intended manners. Student learning in science and technology can be benefited via their conjoint study.  

Yuri [45]3 years ago
4 0
Answer is 
<span>a.Scientific investigation discovers new information that can be used in technological design, and technological design provides new products that can further scientific investigation.

Scientific investigation helps to discover new information, such as greater insight into how the world works. This information can be applied to technological design, illuminating new ways to do things. </span>
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For the following chemical equation.
kotegsom [21]

Sr is the limiting reactant.

Given the reaction equation;

2Sr + O2 (g) → 2SrO

2 moles of Sr reacts with 1 mole of O2

2 moles Sr will react with x mole of O2

x = 2 ×1/2

x = 1 mole of O2

Since we have more O2 than required, it is the reactant in excess, hence Sr is the limiting reactant.

Learn more: brainly.com/question/14225536

5 0
2 years ago
A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th
kondaur [170]

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = \frac{38.8}{61.2^{2} }

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

0.0104 = \frac{200 - P(NO_{2})  }{[P(NO_{2} )]^{2}}

0.0104[P(NO_{2} )]^{2} + P(NO_{2} ) - 200 = 0

Resolving the second degree equation:

P(NO_{2} ) = \frac{-1+\sqrt{9.32} }{0.0208}

P(NO_{2} ) = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are P(NO_{2} ) = 98.7 MPa and P(N₂O₄) = 101.3 MPa

3 0
3 years ago
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Lisa [10]
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3 years ago
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Answer:

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