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3 years ago

What reaction would take place if hot tugsten was surrounded by air?

1 answer:

8 0

The reaction that would take place if hot tungsten was surrounded by air is that it would react to the oxygen in air to form non conducting tungsten oxide and burns out instantly.

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The word igneous comes from the Latin word for fire. However, the name is inaccurate. The rocks are not created with flames. Ins

Fiesta28 [93]

Instead, they are created when magma cools and hardens —> the temperature contributes as the magma would be at extreme heat while it need a cooler temperature to harden and mold

5 0

3 years ago

Read 2 more answers

The reaction 2HgO (s)→2Hg (I)+O2 (g) has a percent yield of 50%. You want to produce 100 g of Hg.

nevsk [136]

The mass of HgO needed for the reaction is 216 g

The correct answer to the question is Option C. 216 g

We'll begin by calculating the theoretical yield of Hg.

Actual yield of Hg = 100 g

Percentage yield = 50%

Theoretical yield of Hg =?

Theoretical yield = Actual yield / percentage yield

Theoretical yield = 100 / 50%

Theoretical yield of Hg = 200 g

Finally, we shall determine the mass of HgO needed for the reaction.

2HgO → 2Hg + O₂

Molar mass of HgO = 201 + 16 = 217 g/mol

Mass of HgO from the balanced equation = 2 × 217 = 434 g

Molar mass of Hg = 201 g/mol

Mass of Hg from the balanced equation = 2 × 201 = 402 g

From the balanced equation above,

402 g of Hg were produced from 434 g of HgO.

Therefore

200 g of Hg will be produce by = (200 × 434) / 402 = 216 g of HgO.

Thus, 216 g of HgO is needed for the reaction.

Learn more about stoichiometry:

brainly.com/question/24426334

4 0

2 years ago

Read 2 more answers

The carbon-carbon bond length in ethylene is ________ than the carbon-carbon bond length in ethane, and the HCH bond angle in et

tigry1 [53]

Answer:

shorter

longer

Explanation:

The carbon-carbon bond length in ethylene is <u>shorter</u> than the carbon-carbon bond length in ethane, and the HCH bond angle in ethylene is <u>longer</u> the HCH bond angle in ethane.

The objective of this question is to let us understand the concept of Bond Length and Bond angle among the unsaturated aliphatic hydrocarbons (i.e alkanes, alkenes and alkynes).

The variation in bond angles of unsaturated aliphatic hydrocarbons can be explained by two concepts; The valence shell electron pair repulsion (VSEPR) model and hybridization.

The VSEPR model determines the total number of electron pairs surrounding the central atom of a species. The total number of electron pairs consist of the bond pairs and lone pairs. All the electron pairs( lie charge ) will then orient themselves in such a way to minimize the electrostatic repulsion between them.

As the number of the lone pairs increases from zero to 2 ; the bond angles diminish progressively.

However;

Hybridization is the mixing or blending of two or more pure atomic orbitals (s,p and d) to form two or more hybrid atomic orbitals that are identical in shape and energy . e.g sp, sp² , sp³ hybrid orbitals etc .

The shape of the geometry of this compound hence determines their bond angle.

The shape of the geometry of ethane is tetrahedral which is 109.5° in bond angle while that of ethylene is trigonal planar which is 120°.

This is why the HCH bond angle in ethylene is longer the HCH bond angle in ethane .

6 0

4 years ago

How many grams of coffee must evaporate from 350 g of coffee in a 100-g glass cup to cool the coffee and the cup from 95.0°C to

Natali5045456 [20]

Answer:

31.3 g

The answer is higher than the true answer.

Explanation:

By neglecting the heat lost by other processes, the energy conservation states that:

Qcooling + Qevaporate = 0

The cooling process happens without phase change, so the heat can be calculated by:

Qcooling = m*c*ΔT

Where m is the mass, c is the heat capacity (cwater = 4184 J/kg.K), and ΔT is the temperature variation (final - initial).

The evaporate process happen without changing of temperature (pure substance), and the heat can be calculated by:

Qevaporate = m*L

Where m is the mass evaporated and L is the heat of evaporation (2340000 J/kg).

0.350*4184*(45 - 95) + m*2340000 = 0

2340000m = 73220

m = 0.0313 kg

m = 31.3 g

Because of the assumptions made, the real mass is not that was calculated. There'll be changing mass when the coffee is cooling, and there'll be heat loses by other processes because the system is not isolated. Also, the substance is not pure. So, there'll be more factors at the energy equation, thus, the answer is higher than the true answer.

6 0

4 years ago

What is impossible to do with a hypothesis

sveticcg [70]

Eat them and drink then

5 0

3 years ago