Molar mass is the number of moles of a substance that have a mass of 1 gram.

Calculate the empirical formula of a compound with the following

Mashutka [201]

Answer:

The empirical formula for this compound is thus CH2.

Explanation:

4 0

3 years ago

PLZ ANSWER

Taya2010 [7]

Answer:

A. The energy stored in atmospheric carbon dioxide is conserved because it is used to create new forms of energy present in decomposed plants.

Explanation:

In the carbon cycle image, the result of an industry's work releases carbon dioxide into the atmosphere (this is represented by the letter G), this carbon dioxide is stored in the atmosphere (letter C) and then absorbed by plants during the process. of photosynthesis (letter A).

The carbon cycle is constituted by the absorption of carbon dioxide by plants in the photosynthesis process. Half of this absorbed carbon is released into the atmosphere and the other half the vegetable uses to produce sugars (glycoses). By ingesting the plants, the animals ingest together the carbon to their body, being released through respiration or decomposition. Because some fungi and bacteria are responsible for the decomposition of both animals and vegetables, they ingest part of this carbon, releasing it into the atmosphere and soil. In addition to bacteria, the burning process also releases carbon dioxide into the soil and atmosphere. Vegetables, through the breathing process, also absorb carbon dioxide and release oxygen unlike animals.

4 0

3 years ago

You wish to prepare a tape-casting slip containing 50 vol% Al2O3 and 50 vol% polyvinyl butyral (PVB) binder. If the density of A

belka [17]

Answer: The mass of PVB required to produce 1000 grams of tape is 213.4 grams

Explanation:

To calculate the mass of aluminium oxide, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$ .......(1)

For $Al_2O_3$

We are given:

50% (v/v) of $Al_2O_3$

This means that 50 mL of aluminium oxide is present in 100 mL of tape

Calculating the mass of aluminium oxide by using equation 1:

Density of aluminium oxide = $3.98 g/cm^3$

Volume of aluminium oxide = $50mL=50cm^3$ (Conversion factor: $1mL=1cm^3$ )

Putting values in equation 1, we get:

$3.98g/cm^3=\frac{\text{Mass of aluminium oxide}}{50cm^3}\\\\\text{Mass of aluminium oxide}=(3.98g/cm^3\times 50cm^3)=199g$

Mass of aluminium oxide = 199 g

For PVB:

We are given:

50% (v/v) of PVB

This means that 50 mL of PVB is present in 100 mL of tape

Calculating the mass of PVB by using equation 1:

Density of PVB = $1.08 g/cm^3$

Volume of PVB = $50mL=50cm^3$

Putting values in equation 1, we get:

$1.08g/cm^3=\frac{\text{Mass of PVB}}{50cm^3}\\\\\text{Mass of PVB}=(1.08g/cm^3\times 50cm^3)=54g$

Mass of PVB = 54 g

Mass of tape = Mass of aluminium oxide + mass of PVB

Mass of tape = [199 + 54] g = 253 g

To calculate the mass of PVB required to produce 1000 g of tape, we use unitary method:

When 253 grams of tape is made, the mass of PVB required is 54 g

So, when 1000 grams of tape is made, the mass of PVB required will be = $\frac{54}{253}\times 1000=213.4g$

Hence, the mass of PVB required to produce 1000 grams of tape is 213.4 grams

4 0

2 years ago

The heat released by one mole of sugar from a bomb calorimeter experiment is 5648 kJ/mol. The balanced chemical reaction equatio

Bogdan [553]

Answer:

The answer to your question is the letter C) 5648 kJ/mol

Explanation:

Data

C₁₂H₂₂O₁₁ + 12 O₂ ⇒ 12 CO₂ + 11 H₂O

H° C₁₂H₂₂O₁₁ = -2221.8 kJ/mol

H° O₂ = 0 kJ / mol

H° CO₂ = -393.5 kJ/mol

H° H₂O = -285.8 kJ/mol

Formula

ΔH° = ∑H° products - ∑H° reactants

Substitution

ΔH° = 12(-393.5) + 11(-285.8) - (-2221.8) - (0)

ΔH° = -4722 - 3143.8 + 2221.8

Result

ΔH° = -5644 kJ/mol

6 0

3 years ago

Help plz:))) I’ll mark u Brainliest

loris [4]

Answer:

We report an unusual case of mercury vapor poisoning from using a heated tobacco product. The suspect had added grains of mercury into 20 cigarettes in a pack. When a 36-year-old Japanese man inserted one of these cigarettes into the battery powered holder, it was heated to a temperature of 350 °C, and he inhaled vaporized mercury. After using 14 of the cigarettes over 16 h, he noticed he had flu-like symptoms so he visited the hospital. Although no physical abnormalities were revealed, 99 μg/L of mercury was detected in his serum sample. His general condition improved gradually and his whole blood mercury level had decreased to 38 μg/L 5 days later. When the remaining six cigarettes in the pack were examined, many metallic grains weighing a total of 1.57 g were observed. Energy dispersive X-ray fluorescence spectrometry confirmed the grains as elemental mercury. Accordingly, the victim was diagnosed with mercury poisoning. Because the mercury was incorporated into cigarettes, an unusual and novel intoxication occurred through the heating of the tobacco product. Both medical and forensic scientific examination confirmed this event as attempted murder.

Explanation:

3 0

2 years ago