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3 years ago

What is the amplitude of a Node?

Chemistry

1 answer:

erastovalidia [21]3 years ago

3 0

Answer by Mimiwhatsup: An amplitude is a node and an point along a standing wave where the wave has minimum amplitude.

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Why isn't air travel based on hydrogen popular today? A.Hydrogen is very expensive B. Hydrogen is difficult to find C. Hydrogen

Butoxors [25]

The answer is c. hydrogen is flammable and dangerous

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3 years ago

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Complete the following reaction :<br><br>HC≡CH + HBr→?

Morgarella [4.7K]

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3 years ago

1. In apothecaries' measures: 1 scruple = 20 grains, 1 ounce = 480 grains, 1 oz = 28.34 g

laila [671]

Answer:

5.89 g × 10⁶ μg

Explanation:

Step 1: Convert 5.00 scruples to grains

We will use the conversion factor 1 scruple = 20 grains.

5.00 scruple × 20 grain/1 scruple = 100 grain

Step 2: Convert 100 grains to ounces

We will use the conversion factor 1 oz = 480 grains.

100 grain × 1 oz/480 grain = 0.208 oz

Step 3: Convert 0.208 oz to grams

We will use the conversion factor 1 oz = 28.34 g.

0.208 oz × 28.34 g/1 oz = 5.89 g

Step 4: Convert 5.89 g to micrograms

We will use the conversion factor 1 g = 10⁶ μg.

5.89 g × 10⁶ μg/1 g = 5.89 g × 10⁶ μg

4 0

3 years ago

RATE LAW QUESTION !

vivado [14]

In general, we have this rate law express.:

$\mathrm{Rate} = k \cdot [A]^x [B]^y$
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation $\Delta \mathrm{Rate} = \Delta [B]^y$

$3.09 = (3)^y \implies y \approx 1$

To this point, $\mathrm{Rate} = k \cdot [A]^x [B]^1$

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for $\Delta \mathrm{Rate} = \Delta [A]^x$

$4= (2)^x \implies x = 2$

the rate law is

Rate = k·[A]²[B]

6 0

3 years ago

A piece of wire contains 1.52x10^22 atoms of copper. Calculate the miles of copper in the wire

melamori03 [73]

Moles of Copper =

$1.52$ × $10^{22}atoms$ × $\frac{1 mol }{6.022 * 10^{23}atoms}$

= 0.0252 mol Cu

6 0

4 years ago