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expeople1 [14]
2 years ago
12

Which is one way that analyzing ice benefits scientists who study ancient climates? Scientists can analyze frozen volcanic dust

to help predict eruptions. Scientists can drill deep into the ice to collect ice cores. Scientists can use pollen grains in ice to make inferences about the climate area. Scientists can study tree rings in ice to learn more about past climates.
Chemistry
1 answer:
aleksandr82 [10.1K]2 years ago
5 0

Answer:

The correct option is;

Scientists can drill deep into the ice to collect ice cores which contain trapped atmospheric gases

Explanation:

The study of past climates also known as paleoclimatology, is accomplished by acquiring information from proxy data sources which are physical environment characteristics that are preserved through time to remake the conditions of past climate

Past physical environmental characteristics, from which information about ancient climate can be gained are stored in nature's climate variability records including, ice cores, rings in tree stems, fossil pollen, sediments found in the waters of the ocean

The proxy sources provide a means of understanding the conditions of ancient climate before advent of climate measurement.

Therefore, one way that analyzing ice benefits scientists who study ancient climates is that scientists can drill deep into the ice to collect ice cores which contain trapped atmospheric gases from past climates.

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Be sure to answer all parts. one of the most important industrial sources of ethanol is the reaction of steam with ethene derive
lions [1.4K]

Answer: 2.17x10⁻³ atm

Explanation:

First, we must write the balanced chemical equation for the process:

C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g)

The chemical reactions that occur in a closed container can reach a state of <u>chemical equilibrium</u> that is characterized because the concentrations of the reactants and products remain constant over time. The <u>equilibrium constant</u> of a chemical reaction is the value of its reaction quotient in chemical equilibrium.

The equilibrium constant (K) is expressed as <u>the ratio between the molar concentrations (mol/L) of reactants and products.</u> Its value in a chemical reaction depends on the temperature, so it must always be specified.

<u>We will use the the equilibrium constant Kc of the reaction to calculate partial pressure of ethene.</u> The constant Kc for the above reaction is,

Kc = \frac{[C_{2} H_{5}OH]}{[H_{2}O][C_{2} H_{4}]}

According to the law of ideal gases,  

PV = nRT  

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the gas constant (0.082057 atm L / mol K) .

We can use the ideal gas law to determine the molar concentrations ([x] = n / V) from the gas pressures of ethanol and water, assuming that all gases involved behave as ideal gases. In this way,

PV = nRT → P = (n/V) RT → P = [x] RT → [x] = P / RT

So,  

[C_{2} H_{5}OH] = \frac{200 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 4.06 \frac{mol}{L}

[H_{2}O] = \frac{400 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 8.12 \frac{mol}{L}

So, the molar concentration of ethene (C₂H₄) will be,

[C_{2} H_{4}] = \frac{[C_{2} H_{5}OH]}{[H_{2}O] x Kc} = \frac{4.06 \frac{mol}{L} }{8.12 \frac{mol}{L}x9.00 x 10^{3} \frac{L}{mol} } = 5.56 x 10^{-5}\frac{mol}{L}

Then, according to the law of ideal gases,

P_{C_{2} H_{4}} = [C_{2} H_{4}]RT = 5.56 x 10^{-5} \frac{mol}{L}  x 0.082057 \frac{atm L}{mol K} x 600 K = 2.17x10^{-3} atm

So, when the partial pressure of ethanol is 200 atm and the partial pressure of water is 400 atm, the partial pressure of ethene at 600 K is 2.17x10⁻³ atm.

7 0
2 years ago
Balance the following redox reaction occurring in an acidic solution. The coefficient of Cr2O72−(aq) is given. Enter the coeffic
ollegr [7]

Answer:

Cr₂O₇²⁻ (aq) + 6Ti³⁺ (aq) + 2 H⁺(aq) → 2 Cr³⁺ (aq) + 6TiO²⁺(aq) + H2O(l)

Explanation:

Cr₂O₇²⁻ (aq) + Ti³⁺ (aq) +  H⁺ (aq)  → Cr³⁺ (aq) + TiO²⁺ (aq) +  H₂O(l)

This is the reaction, without stoichiometry.

We have to notice the oxidation number of each element.

In dicromate, Cr acts with +6 and we have Cr³⁺, so oxidation number has decreased. .- REDUCTION

Ti³⁺ acts with +3, and in TiO²⁺ acts with +4 so oxidation number has increased.  .- OXIDATION

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O

To change 6+ to 3+, Cr had to lose 3 e⁻, but as we have two Cr, it has lost 6e⁻. As we have 7 Oxygens in reactant side, we have to add water, as the same amount of oxygens atoms we have, in products side. Finally, we have to add protons in reactant side, to ballance the H.

H₂O  +  Ti³⁺  →  TiO²⁺ + 1e⁻ + 2H⁺

Titanium has to win 1 e⁻ to change 3+ to 4+. We had to add 1 water in reactant, and 2H⁺ in products, to get all the half reaction ballanced.

Now we have to ballance the electrons, so we can cancel them.

(14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O) .1

(H₂O  +  Ti³⁺  →  TiO²⁺ + 1e⁻ + 2H⁺) .6

Wre multiply, second half reaction .6

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O

6H₂O  +  6Ti³⁺  →  6TiO²⁺ + 6e⁻ + 12H⁺

Now we can sum, the half reactions:

14 H⁺ + Cr₂O₇²⁻ + 6e⁻  + 6H₂O  +  6Ti³⁺  → 6TiO²⁺ + 6e⁻ + 12H⁺ + 2Cr³⁺ + 7H₂O

Electrons are cancelled and we can also operate with water and protons

7H₂O - 6H₂O = H₂O

14 H⁺ - 12H⁺ = 2H⁺

The final ballanced equation is:

2H⁺ + Cr₂O₇²⁻ + 6Ti³⁺  → 6TiO²⁺  + 2Cr³⁺ + H₂O

8 0
2 years ago
A student sets up the following equation to convert a measurement. (The stands for a number the student is going to calculate.)
Aliun [14]

The question is incomplete, here is the complete question:

A student sets up the following equation to convert a measurement. (The stands for a number the student is going to calculate.) Fill in the missing part of this equation.

23.Pa.cm^3=?kPa.m^3

<u>Answer:</u> The measurement after converting is 23\times 10^{-9}kPa.m^3

<u>Explanation:</u>

We are given:

A quantity having value 23.Pa.cm^3

To convert this into kPa.m^3, we need to use the conversion factors:

1 kPa = 1000 Pa

1m^3=10^6cm^3

Converting the quantity into kPa.m^3, we get:

\Rightarrow 23.Pa.cm^3\times (\frac{1kPa}{1000Pa})\times (\frac{1m^3}{10^6cm^3})\\\\\Rightarrow 23\times 10^{-9}kPa.m^3

Hence, the measurement after converting is 23\times 10^{-9}kPa.m^3

6 0
2 years ago
Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given
professor190 [17]

Answer:  -227 kJ

Explanation:

The balanced chemical reaction is,

C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]

-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]

\Delta H_{C_2H_2}=-227kJ

Therefore, the enthalpy change for C_2H_2 is -227 kJ.

7 0
2 years ago
A container holds 15.0 g of phosphorous gas at a pressure of 2.0 atm and a temperature of 20.0 Celsius. What is the density of t
erik [133]
<span>Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. We calculate as follows:

PV = nRT
PV = mRT/ Molar mass
m/V = P(molar mass)/RT
Density = P(molar mass)/RT 
Density = 2.0 ( 30.97 ) / 0.08206 ( 20 + 273.15) = 2.57 g/L <----First option</span>
5 0
2 years ago
Read 2 more answers
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