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3 years ago

How are computers used in producing weather forecasts ?

1 answer:

7 0

Supercomputers are used in weather forecasting as they need to store data from various instruments including radars, satellites, and so on.

Explanation:

Weather forecasting uses supercomputers. They employ arithmetical models that can predict and forecast the weather over time.
The present condition of the atmosphere is first taken. Various weather parameter variables like temperature, pressure, wind speed, and so on are collected and stored.
The interpolation and and complex values to analyze are hence done by the models in a super computer.

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When 47.1 J of heat is added to 14.0 g of a liquid, its temperature rises by 1.80 ∘C. What is the heat capacity of the liquid?

Alja [10]

Answer:

$\boxed {\boxed {\sf 1.87 \J/g \textdegree C}}$

Explanation:

We are asked to find the specific heat capacity of a liquid. We are given the heat added, the mass, and the change in temperature, so we will use the following formula.

$q= mc\Delta T$

The heat added (q) is 47.1 Joules. The mass (m) of the liquid is 14.0 grams. The specific heat (c) is unknown. The change in temperature (ΔT) is 1.80 °C.

q= 47.1 J
m= 14.0 g
ΔT= 1.80 °C

Substitute these values into the formula.

$47.1 \ J = (14.0 \ g) * c * (1.80 \textdegree C)$

Multiply the 2 numbers in parentheses on the right side of the equation.

$47.1 \ J = (14.0 \ g * 1.80 \textdegree C)*c$

$47.1 \ J = (25.2 \ g*\textdegree C) *c$

We are solving for the heat capacity of the liquid, so we must isolate the variable c. It is being multiplied by 25.2 grams * degrees Celsius. The inverse operation of multiplication is division, so we divide both sides of the equation by (25.2 g * °C).

$\frac {47.1 \ J}{(25.2 g *\textdegree C)} = \frac {(25.2 g *\textdegree C)*c}{{(25.2 g *\textdegree C)}}$

$\frac {47.1 \ J}{(25.2 g *\textdegree C)} =c$

$1.869047619 \ J/g *\textdegree C = c$

The original measurements of heat, mass, and temperature all have 3 significant figures, so our answer must have the same. For the number we found that is the hundredth place. The 9 in the thousandth place to the right tells us to round the 6 up to a 7.

$1.87 \ J/ g * \textdegree C =c$

The heat capacity of the liquid is approximately 1.87 J/g°C.

3 0

3 years ago

A beaker with 1.60×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and c

stich3 [128]

Answer:

The pH will change 0.16 ( from 5.00 to 4.84)

Explanation:

Step 1: Data given

volume of acetic acid buffer = 160 mL

The total molarity of acid and conjugate base in this buffer is 0.100 M

A student adds 7.10 mL of a 0.460 M HCl solution to the beaker.

The pKa of acetic acid is 4.740

pH = 5.00

Step 2: Calculate concentration of acid

Consider x = concentration acid

Consider y = concentration conjugate base

x + y = 0.100

5.00 = 4.740 + log y/x

5.00 - 4.740 = log y/x

0.26 = log y/x

10^0.26 =1.82 = y/x

1.82 x = y

Since x+y = 0.100

x + 1.82 x = 0.100

2.82 x = 0.100

x =0.0355 M = concentration acid

Step 3: Calculate concentration of conjugate base

y = 0.100 - x

0.100 - 0.0355 =0.0645 M= concentration conjugate base

Step 4: Calculate moles of acid

Moles = volume * molarity

moles acid = 0.160 L * 0.0355 M= 0.00568 moles

Step 5: Calculate moles of conjugate base

moles conjugate base = 0.0645 M * 0.160 L=0.01032 moles

Step 6: Calculate moles HCl

moles HCl = 7.10 * 10^-3 L * 0.460 M=0.003266 moles

Step 7: Calculate new moles

A- + H+ = HA

moles conjugate base = 0.01032 - 0.003266 =0.007054 moles

moles acid = 0.00568 + 0.003266=0.008946 moles

Step 8: Calculate the total volume

total volume = 160 + 7.10 = 167.1 mL = 0.1671 L

Step 9: Calculate the concentration of the acid

concentration acid = 0.008946/ 0.1671 =0.0535 M

Step 10: Calculate the concentration of conjugate base

concentration conjugate base = 0.007054/ 0.1671 =0.0422 M

Step 11: Calculate the pH

pH = 4.740 + log 0.0535/ 0.0422=4.84

change pH = 5.00 - 4.84=0.16

The pH will change 0.16

5 0

3 years ago

Explain how low atomspheric oxygen levels affected alligators growth in terms of celluar respiration

Paraphin [41]

Answer:

Answer: Hyperoxic alligators show significantly lower breathing rates than their siblings in normoxic and hypoxic groups. Conversely, each breath supports greater oxygen consumption in hyperoxic animals than in other groups. Statistical significance between groups was calculated by ANOVA with post hoc Tukey-Kramer

3 0

3 years ago

Which of the following substances is most likely to be a liquid at room temperature?

nikitadnepr [17]

Answer:

The correct answer is d.hydrogen peroxide H₂O₂H₂O₂

Explanation:

Substances can be found in nature in different <em>aggregation states. </em>

Agreggation states can be liquid, gas or solid.

The problem asks about which substance can be found in room temperature as <em>liquid.</em> Each subtances has different physical and chemical properties that determines in which state you can find them at room temperature.

Hydrogen peroxide is the only substance listed that is in liquid state at room temperature, all other substances are in gas state at room temperature.

3 0

3 years ago

Calculate ΔrG∘ at 298 K for the following reactions.CO(g)+H2O(g)→H2(g)+CO2(g)2-Predict the effect on ΔrG∘ of lowering the temper

KonstantinChe [14]

Answer:

1) ΔG°r(298 K) = - 28.619 KJ/mol

2) ΔG°r will decrease with decreasing temperature

Explanation:

CO(g) + H2O(g) → H2(g) + CO2(g)

1) ΔG°r = ∑νiΔG°f,i

⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)

from literature, T = 298 K:

∴ ΔG°CO2(g) = - 394.359 KJ/mol

∴ ΔG°CO(g) = - 137.152 KJ/mol

∴ ΔG°H2(g) = 0 KJ/mol........pure substance

∴ ΔG°H2O(g) = - 228.588 KJ/mol

⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )

⇒ ΔG°r(298 K) = - 28.619 KJ/mol

2) K = e∧(-ΔG°/RT)

∴ R = 8.314 E-3 KJ/K.mol

∴ T = 298 K

⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6

⇒ ΔG°r = - RTLnK

If T (↓) ⇒ ΔG°r (↓)

assuming T = 200 K

⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)

⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol

6 0

4 years ago