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Rzqust [24]
2 years ago
9

Noah is conducting a science experiment and needs to analyze iron and sulfur. Which of the following will he compare to determin

e the amounts of each substance?
movement of the atoms in each substance
movement of the atoms in each substance

molecular mass of each substance
molecular mass of each substance

mass of each substance
mass of each substance

vibration of the atoms in each substance
vibration of the atoms in each substance
Chemistry
1 answer:
qaws [65]2 years ago
7 0

Answer:

molecular mass of each substance

molecular mass of each substance

Explanation:

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Answer: Chemical Change

Explanation:

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3 years ago
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The Great Plains are also referred to as____.
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The interior plains

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2.92 x 10^-15/<br> (5.6 x 10^-3) (4.16 x 10^9)
s2008m [1.1K]

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3645.46

Explanation:

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3 years ago
How many grams of iron are in 350 mg of iron?-
Mamont248 [21]

Answer:

0.350 g of iron

Explanation:

Step 1: Given data

Mass of iron (m): 350 mg

Step 2: Convert the mass of iron to milligrams

In order to convert the mass of iron from grams to milligrams we need a conversion factor. In this case, the conversion factor is 1 g = 1,000 mg.

350 mg Fe × 1 g Fe/1,000 mg Fe = 0.350 g Fe

350 milligrams of iron is equal to 0.350 grams of iron. We conserve the 3 significante figures of the original data.

3 0
3 years ago
What volume of methane gas at 237 K and 101.33 kPa do you have when the volume is decreased to 0.50 L, with a temperature of 300
Alexxandr [17]

Answer: A volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

Explanation:

Given: T_{1} = 237 K,   P_{1} = 101.33 kPa,      V_{1} = ?

T_{2} = 300 K,      P_{2} = 151.99 kPa,        V_{2} = 0.50 L

Formula used is as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{101.33 kPa \times V_{1}}{237 K} = \frac{151.99 kPa \times 0.50 L}{300 K}\\V_{1} = 0.592 L

Thus, we can conclude that a volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

8 0
3 years ago
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