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erik [133]
2 years ago
6

If 11.87 grams of sand (SiO2) were contained in the mixture, how many atoms of oxygen were in the mixture?

Chemistry
1 answer:
ozzi2 years ago
6 0

Answer:

2.38x10²³ atoms oxygen

Explanation:

To solve this question we need to convert the mass of sand to moles using its molar mass (Molar mass SiO₂ = 60.08g/mol). Twice these moles are the moles of Oxygen and using Avogadro's number we can find the amount of atoms of Oxygen in the mixture:

<em>Moles SiO₂:</em>

11.87g * (1mol / 60.08g) = 0.1976moles SiO₂

<em>Moles Oxygen:</em>

0.1976moles SiO₂* 2 = 0.3951moles oxygen

<em>Atoms oxygen:</em>

0.3951moles oxygen * (6.022x10²³atoms / 1mol O₂) =

<h3>2.38x10²³ atoms oxygen</h3>
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g100num [7]

Answer:

Transitional metals are good conductors of both heat and electricity.

But since there aren't any. It could be Calcium.

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3 years ago
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A) 24 g<br> B) 36 g<br> C) 48 g<br> D) 60 g
Hitman42 [59]

Answer:

I'm not sure if this right because i've only taken one year of Chemistry but the answer I got was

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5 0
2 years ago
A 60.0 g block of iron that has an initial temperature of 250. °C and 60.0 g bloc of gold that has an initial temperature of 45.
Maslowich

Answer:

The final temperature at the equilibrium is 204.6 °C

Explanation:

Step 1: Data given

Mass of iron = 60.0 grams

Initial temperature = 250 °C

Mass of gold = 60.0 grams

Initial temperature of gold = 45.0 °C

The specific heat capacity of iron = 0.449 J/g•°C

The specific heat capacity of gold = 0.128 J/g•°C.

Step 2: Calculate the final temperature at the equilibrium

Heat lost = Heat gained

Qlost = -Qgained

Qiron = -Qgold

Q=m*c*ΔT

m(iron) * c(iron) *ΔT(iron) = -m(gold) * c(gold) *ΔT(gold)

⇒with m(iron) = the mass of iron = 60.0 grams

⇒with c(iron) = the specific heat of iron = 0.449 J/g°C

⇒with ΔT(iron)= the change of temperature of iron = T2 - T1 = T2 - 250.0°C

⇒with m(gold) = the mass of gold= 60.0 grams

⇒with c(gold) = the specific heat of gold = 0.128 J/g°C

⇒with ΔT(gold) = the change of temperature of gold = T2 - 45.0 °C

60.0 *0.449 * (T2 - 250.0) = -60.0 * 0.128 * (T2 - 45.0 )

26.94 * (T2 - 250.0) = -7.68 * (T2 - 45.0)

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34.62T2 = 7080.6

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The final temperature at the equilibrium is 204.6 °C

5 0
3 years ago
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Answer:

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Explanation:

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