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3 years ago

What do AIA, SAG, AFP and AMA all stand for? Select four options.

Chemistry

1 answer:

astra-53 [7]3 years ago

3 0

Answer:

ALA.American Institute of Architects

SAG.Screen Actor Guild

AFP.Association For Finance Professional

AMA.American Medical Association

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If the copper is drawn into wire whose diameter is 9.50 mm, how many feet of copper can be obtained from the ingot? The density

ankoles [38]

Answer:

The length of the wire = 352.66 feet.

Explanation:

A copper refinery produces a copper ingot weighing 150 lb. If the copper is drawn into wire whose diameter is 9.50 mm, how many feet of copper can be obtained from the ingot? The density of copper is 8.94 g/cm3. (Assume that the wire is a cylinder whose volume is V = πr2h, where r is the radius and h is its height or length.)

Step 1: Convert lb to kg

150 lb = 68.0389 kg

Step 2: Calculate volume of copper

Volume = mass / density

Volume = 68038.9 grams / 8.94 g/cm³

Volume = 7610.6 cm³ Cu

Step 3: Calculate length of wire

The diameter of the wire is 9.50 mm, so the radius is half of that (4.75 mm), or 0.475 cm.

The total "volume" of the wire is πr²h = (π)*(0.475 cm)²(h) = 0.708h = 7610 cm^3

7610 = 0.708h

h = 10749 cm = length of wire

The length of the wire = 352.66 feet.

7 0

4 years ago

6. The nuclear reaction where Si-30 becomes P-30 is

Arisa [49]

Answer:c

Explanation:

8 0

3 years ago

There are three naturally occurring isotopes of the hypothetical element hilarium 45Hi, 46Hi, and 48Hi. The percentages of these

den301095 [7]

Answer:

46.761g/mol

Explanation:

Given parameters:

Element = Hilarium , Hi

Isotopes: Hi- 45, Hi-46 and Hi- 48

Natural abundance of Hi-45 = 18.3%

Hi-46 = 34.5%

Hi-48 = 47.2%

Unknown:

Atomic weight of naturally occurring Hilarium = ?

Solution:

Isotopes have been studied extensively by mass spectrometry. The method is used to determine the proportion/percentage/fraction by which each of the isotopes of an element occurs in nature. The proportion is called geonormal abundance. From this we can calculate the atomic weight of an element.

We can use the expression below to find this value:

Atomic weight = m₄₅α₄₅ + m₄₆α₄₆ + m₄₈α₄₈

m is the atomic mass of each isotope and α is the abundance

Atomic weight = (45 x $\frac{18.3}{100}$ ) + (46 x $\frac{34.5}{100}$ ) + (48 x $\frac{47.2}{100}$ )