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3 years ago

Calculate the change in temperature that occurs when 3.50 x 10^-2kg of copper heated by 635 J of energy. Ccu = 0.385 j/gxC

Chemistry

2 answers:

vovikov84 [41]3 years ago

6 0

Answer:

The change in T° is 47.1 °C

Explanation:

Calorimetry formula to solve this:

Q = m . C . ΔT

We replace the data gien:

635 J = 3.50×10⁻² kg . 0.385 J/g°C . ΔT

In units of C, we have g and the mass (m) is in kg. So let's convert it from kg to g → 3.50×10⁻² kg . 1000 g / 1kg = 35 g. Now we can determine the ΔT:

635 J = 35 g . 0.385 J/g°C . ΔT

635 J / 35 g . 0.385 J/g°C = ΔT

47.1°C = ΔT

drek231 [11]3 years ago

6 0

Answer:

The change in temperature is 47.12 °C

Explanation:

Step 1: Data given

Mass of copper = 3.50 * 10^-2 kg = 35 grams

Energy = 635 J

Specific heat of copper = 0.385 J/g°C

Step 2: Calculate change in temperature

Q = m*c*ΔT

⇒ Q = the energy required = 635 J

⇒ m = the mass of copper = 35 grams

⇒ c = the specific heat of copper= 0.385 J/g°C

⇒ ΔT = Change in temperature = TO BE DETERMINED

635 J = 35g * 0.385 J/g°C * ΔT

ΔT = 635 / (35*0.385)

ΔT = 47.12 °C

The change in temperature is 47.12 °C

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onsider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 L of a dye solution with a c

Juli2301 [7.4K]

Answer:

Explanation:

From the given information;

Let Q(t) = mass of dye in the tank as a function of time

The mass in the tank = 200 L × (1g/L) = 200 g

Using the law of mass conservation;

$\dfrac{dQ}{dt} = \text{(Rate of incoming mass)- (rate of outgoing mass)}$

$\dfrac{dQ}{dt} = (2 L/min ) (0 \ g/L) - (2 L/min ) (\dfrac{Q}{200}g/L)$

$Q' = \dfrac{-Q}{1000}$

Q(0) = 200

By finding the solution to the ODE using the method of separation of variables;

$\dfrac{Q'}{Q} = -0.01$

$Q(t) = Ce^{-0.01t}$

Using the initial condition;

200 = Q(0) = C

$Q(t) = 200e^{-0.01t}$

1% of 200g = 2g of dye solution

∴

$2 = 200e^{-0.01t}$

$e^{-0.01t}=0.01$

$t =\dfrac{ In(0.01}{-0.01}$

t = 460.5 hours

4 0

3 years ago

Which of the following are Arrhenius Bases? <br> A) CH3OH<br> B) CH3COOH<br> C) HOH<br> D) H2NNH2

Viktor [21]

<h3>Answer:</h3>

H₂O

Explanation:

According to Arrhenius Concept of Acid and Base, "Acid is any specie which when dissolved in water ionizes to produce H⁺ ions".

Examples:

HCl → H⁺ + Cl⁻

H₂SO₄ → 2 H⁺ + SO₄²⁻

While, "Bases are those species which when dissolved in water produces Hydroxyl Ions (OH⁻)".

Examples:

NaOH → Na⁺ + OH⁻

Mg(OH)₂ → Mg²⁺ + 2 OH⁻

Given Options:

CH₃OH:

Methanol when dissolved in water does not dissociate as follow,

CH₃OH → H₃C⁺ + OH⁻

Hence, it does not behave as Arrhenius Base.

CH₃COOH:

Acetic acid when dissolved in water produces H⁺ ions and Acetate ions i.e.

CH₃COOH → CH₃COO⁻ + H⁺

Therefore, it can act as Arrhenius Acid instead of Arrhenius Base.

H₂O:

Water when dissolved in water dissociates as,

H₂O → H⁺ + OH⁻

As it is producing both H⁺ and OH⁻ ions therefore, it can act as both Arrhenius Base and Arrhenius Acid.

H₂NNH₂:

Hydrazine when dissolved in water can neither produce OH⁻ ions nor H⁺ ions hence, it is neither Arrhenius Base nor Arrhenius Acid,

4 0

3 years ago

3 types of spheres of the earth and activities that occur in each

slamgirl [31]

The Geosphere - meaning 'ground', types of rocks, like mountain rocks, as wells as liquid rock in the mantle below us, the minerals and metals of the outer and inner core, all the continents on the oceanic floor.

The Hydrosphere - meaning 'water', any type of water, like rivers, ocean, lakes, streams, groundwater, polar ice caps, glaciers and moisture in the air like the snow and rain.

The Biosphere - meaning 'life', all the living organisms on the planet, like animals, plants, bacteria, fungi, all of these are on the earth or no deeper then 10 feet into the ground or about 600 feet above.

---

There is actually another one and it's

The Atmosphere - meaning 'air', any type of air that surrounds the earth and that floats in earth, like oxygen the one we breath and plants give out, carbon dioxide which plants breath in.

You chose which one is the most important one !!!!

5 0

3 years ago

A chemist has one solution that is 20% salt and a second solution that is 45% salt. How many liters of each should be used in or

Free_Kalibri [48]

Answer:

8 liters of the first salt solution and 12 liters of the second salt solution is needed to get 20L of a solution that is 30% salt

Explanation:

Let x represent the liters of the first salt solution

Let y represent the liters of the second salt solution

x + y= 20......equation 1

x= 20-y

The first solution is 20% salt and the second solution is 45% salt

20/100x + 45/100 y= 30/100 × 20

0.2x + 0.45y= 0.3×20

0.2x + 0.45y= 6............equation 2

Substitute 20-y for x in equation 2

0.2(20-y) + 0.45y= 6

4-0.2y+ 0.45y= 6

4 + 0.25y= 6

0.25y= 6-4

0.25y= 2

y= 2/0.25

y= 8

Substitute y for 8 in equation 1

x + y= 20

x + 8=20

x= 20-8

x= 12

Hence 8 liters of the first salt solution and 12 liters of the second salt solution is needed to get 20L of a solution that is 30% salt.

4 0

2 years ago

A treatment plant uses a CMFR as the reactor for the removal of manganese via an oxidation reaction by the addition of potassium

Alinara [238K]

Answer:

the effluent concentration is 0.06236 mg/L

Explanation:

Given that;

treatment capacity $V_{0}$ = 3,800 m³/d = ( 3,800 × 86.4) = 43.98 L/sec

reactor's volume V = 45 m³ = (45 × 1000) = 45,000 L

reaction rate constant K = 0.0125 s⁻¹

influent manganese concentration $CA_{0}$ = 0.86 mg/L

- $r_{A}$ = $KC_{A}$

Now, performance equation for CSTR is expressed as follows;

$\frac{V}{V_{0} }$ = $\frac{CA_{0} -CA _{} }{-r_{A} }$

$\frac{V}{V_{0} }$ = $\frac{CA_{0} -CA _{} }{KC_{A} }$

So we substitute

45000L / 43.98 L/sec = ( 0.86 mg/L - CA) / 0.0125 CA

we cross multiply

562.5CA = 37.8228 - 43.98CA

562.5CA + 43.98CA = 37.8228

606.48CA = 37.8228

CA = 37.8228 / 606.48

CA = 0.06236 mg/L

Therefore, the effluent concentration is 0.06236 mg/L

3 0

2 years ago