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2 years ago

Which of the following remains would be most likely only be found as a compression fossil?

Chemistry

2 answers:

Anit [1.1K]2 years ago

6 0

Answer:

Leaf

Explanation:

Because it is a plant

attashe74 [19]2 years ago

3 0

Answer:

A Leaf

Explanation:

Compressions are the way that leaves are fossilized as the leaves are pressed firmly between hard layers of rock.

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Determine the potential deaths resulting from the following exposure to chlorine:

jekas [21]

Answer:

For each scenario as following:

A. 3 Potential deaths by chlorine exposure

B. 1 Potential deaths by chlorine exposure

C. 3 Potential deaths by chlorine exposure

Explanation:

According to Freitag, 1941 Chlorine exposure can be lethal at the concentration of 34-51 ppm in a time of 1h-1.5h. The answers are based on his reference.

6 0

3 years ago

I NEED HELP PLEASE !!!!!

kolbaska11 [484]

Answer: jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj

Explanation:

4 0

2 years ago

The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released f

frutty [35]

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 4.60 atm = 7.59 \times 10^{-3} M$

We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.

$\frac{7.59 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.367 g$

Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 1.28 atm = 2.11 \times 10^{-3} M$

$\frac{2.11 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.102 g$

The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

$m = 0.367 g - 0.102 g = 0.265 g$

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

Learn more: brainly.com/question/18987224

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.

5 0

2 years ago

PLS HELP QUICK ALOTTT OF POINTS

timofeeve [1]

Answer:

$\boxed {\boxed {\sf 0.80 \ mol\ F}}$

Explanation:

We are asked to find how many moles are in 4.8 × 10²³ fluorine atoms. We convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of fluorine.

We will convert using dimensional analysis and set up a ratio using Avogadro's Number.

$\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}$

We are converting 4.8 × 10²³ fluorine atoms to moles, so we multiply the ratio by this number.

$4.8 \times 10^{23} \ atoms \ F *\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}$

Flip the ratio so the units of atoms of fluorine cancel each other out.

$4.8 \times 10^{23} \ atoms \ F *\frac { 1 \ mol \ F}{6.022 \times 10^{23} \ atoms \ F}$

$4.8 \times 10^{23} *\frac { 1 \ mol \ F}{6.022 \times 10^{23} }$

Condense into 1 fraction.

$\frac { 4.8 \times 10^{23} }{6.022 \times 10^{23} } \ mol \ F$

Divide.

$0.7970773829 \ mol \ F$

The original measurement of atoms has 2 significant figures, so our answer must have the same. For the number we found, that is the hundredths place. The 7 in the thousandths tells us to round the 9 in the hundredths place up to a 0. Then, we also have to round the 7 in the tenths place up to an 8.

$0.80 \ mol \ F$

4.8 × 10²³ fluorine atoms are equal to 0.80 moles of fluorine.

6 0

3 years ago

When 0.513 g of biphenyl (c12h10 undergoes combustion in a bomb calorimeter, the temperature rises from 26.3 ?c to 29.7 ?c?

olasank [31]

When 0.514 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.8 C to 29.4 C. Find ⌂E rxn for the combustion of biphenyl in kJ/mol biphenyl. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/ C.

The answer is - 6.30 * 10^3 kJ/mol

3 0

3 years ago