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1 year ago

Explain how layers that form in ice are similar to tree rings?

2 answers:

4 0

Layers of ice and tree rings are similar in that they can both record or store data from past environmental conditions. The study of tree rings is known as dendrochronology and it involves counting and studying individual rings in the tree. Each ring represents a year or specific growing period. If the ring is smaller it means there might have been a drought or nutrients were low. A larger than average ring would represent an unusually favourable growing season. Layers of ice in an ice sheet trap information that can tell about past conditions such as climate. Just like the tree rings specific sections represent a specific period of time. Air bubbles trapped in ice provides information on past climates.

Ber [7]1 year ago

4 0

Answer:

<h2>Answer is </h2><h2></h2>

Each layer found in ice represents one year of snow fall. Each tree ring represents one year of growth. Both ice layers and tree rings, based on their size and make up, can tell us about the climate for that given year.

(When writing down the answer please change some words, Thank You)

Explanation:

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En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello

MAXImum [283]

Answer:

a) $m_{CH_4}=2630kg$

b) 1657 €

Explanation:

Hola,

a) En este problema, vamos a considerar el millón de litros de agua anuales, ya que con ellos podemos calcular el calor requerido para dicho calentamiento, sabiendo que la densidad del agua es de 1 kg/L:

$Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ$

Luego, usamos la entalpía de combustión del metano para calcular su requerimiento en kilogramos, sabiendo que la energía ganada por el agua, es perdida por el metano:

$Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}$

$m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg$

b) En este caso, consideramos que a condiciones normales de 1 bar y 273 K, 1 metro cúbico de metano cuesta 0,45 €, con esto, calculamos las moles de metano a dichas condiciones:

$n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol$

Con ello, los kilogramos de metano que cuestan 0,45 €:

$44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4$

Luego, aplicamos la regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

3 0

2 years ago

Consider the equation: 2NO2(g) N2O4(g). Using ONLY the information given by the equation which of the following changes would in

Reika [66]

I believe the correct answer is the first option. To increase the molar concentration of the product N2O4, you should increase the pressure of the system. You cannot determine the effect of changing the temperature since we cannot tell whether it is an endothermic or an exothermic reaction. Also, decreasing the number of NO2 would not increase the product rather it would shift the equilibrium to the left forming more reactants. The only parameter we can change would be the pressure. And, since NO2 takes up more space than the product increasing the pressure would allow the reactant to collide more forming the product.

7 0

1 year ago

Read 2 more answers

93.2 mL of a 2.03 M potassium fluoride (KF) solution

Marrrta [24]

Answer:

1.98 M

Explanation:

Given data

Initial volume (V₁): 93.2 mL

Initial concentration (C₁): 2.03 M

Volume of water added: 3.92 L

Step 1: Convert V₁ to liters

We will use the relationship 1 L = 1000 mL.

$93.2mL \times \frac{1L}{1000mL} = 0.0932 L$

Step 2: Calculate the final volume (V₂)

The final volume is the sum of the initial volume and the volume of water.

$V_2 = 0.0932L + 3.92 L = 4.01L$

Step 3: Calculate the final concentration (C₂)

We will use the dilution rule.

$C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{2.03 M \times 3.92L}{4.01L} = 1.98 M$

3 0

1 year ago

Read 2 more answers

By mistake, a quart of oil was dumped into a swimming pool that measures 25.0 m by 30.0 m. The density of the oil was 0.750 g/cm

kondor19780726 [428]

The oil slick thick = 1.256 x 10⁻⁴ cm

<h3>Further explanation</h3>

Volume is a derivative quantity derived from the length of the principal

The unit of volume can be expressed in liters or milliliters or cubic meters

The conversion is

1 cc = 1 cm3

1 dm = 1 Liter

1 L = 1.06 quart

<em>so for 1 quart = 0.943 L</em>

$\tt 0.943~L=0.943\times 10^{-3}m^3$

Volume of oil dumped = volume of swimming pool

$\tt 0.943\times 10^{-3}~m^3=25\times 30\times h(h=thick)\\\\h=\dfrac{0.943\times 10^{-3}}{750~m^2}=1.257\times 10^{-6}~m=\boxed{\bold{1.256\times 10^{-4}~cm}}$

3 0

1 year ago

At which temperature do the molecules of an ideal gas have 3 times the kinetic energy they have at 32of?

algol [13]

Answer:

820 K

Explanation:

As per Boltzman equation, <em>kinetic energy (KE)</em> is in direct relation to the <em>temperature</em>, measured in absolute scale Kelvin.

KE α T.

Then, <em>the temperature at which the molecules of an ideal gas have 3 times the kinetic energy they have at any given temperature will be </em><em>3 times</em><em> such temperature.</em>

So, you must just convert the given temperature, 32°F, to kelvin scale.

You can do that in two stages.

First, convert 32°F to °C. Since, 32°F is the freezing temperature of water, you may remember that is 0°C. You can also use the conversion formula: T (°C) = [T (°F) - 32] / 1.80

Second, convert 0°C to kelvin:

T (K) = T(°C) + 273.15 K= 273.15 K

Then, <u>3 times</u> gives you: 3 × 273.15 K = 819.45 K

Since, 32°F has two significant figures, you must report your answer with the same number of significan figures. That is 820 K.

7 0

1 year ago