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3 years ago

PLEASE HELP ASAP!!!!

Chemistry

1 answer:

Vsevolod [243]3 years ago

4 0

Answer:

The answer to your question is C = 0.037 cal/g°C

Explanation:

Data

mass of the wire = m = 237 g

temperature 1 = T1 = 25°C

temperature 2 = T2 = 107°C

Heat = Q = 722 cal

Specific heat = C

Process

1.- Write the formula to find the specific heat

Q = mC(T2 - T1)

-Solve for C

C = Q / m(T2 - T1)

2.- Substitution

C = 722 / 237(107 - 25)

3.- Simplification

C = 722 / 237(82)

C = 722 / 19434

4.- Result

C = 0.037 cal/g°C

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julia-pushkina [17]

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3 years ago

It's 32°F (0°C) outside under normal atmospheric conditions (1 atm) at a stunt

Korolek [52]

The volume of the balloon is approximately 2652 liters.

<h3>How to determine the volume occupied by the gas in a balloon </h3>

Let suppose that <em>flammable</em> hydrogen behaves ideally. GIven the molar mass ( $M$ ), in kilograms per kilomole, and mass of the gas ( $m$ ), in kilograms. The volume occupied by the gas ( $V$ ), in cubic centimeters, is found by the equation of state for <em>ideal</em> gases:

$V = \frac{m\cdot R_{u}\cdot T}{P\cdot M}$ (1)

Where:

$R_{u}$ - Ideal gas constant, in kilopascal-cubic meters per kilomole-Kelvin.
$T$ - Temperature, in Kelvin
$P$ - Pressure, in kilopascals

If we know that $m = 0.239\,kg$ , $R_{u} = 8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}$ , $T = 273.15\,K$ , $P = 101.325\,kPa$ and $M = 2.02\,\frac{kg}{kmol}$ , then the volume of the balloon is:

$V = \frac{(0.239\,kg)\cdot \left(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)\cdot (273.15\,K)}{\left(101.325\,kPa\right)\cdot \left(2.02\,\frac{kg}{kmol} \right)}$

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The volume of the balloon is approximately 2652 liters. $\blacksquare$

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1 year ago

if the lightbulb receives 100 J of electrical energy, and gives off 75 energy, how much heat (thermal energy away from the light

oksano4ka [1.4K]

Answer:

Amount of heat energy released by light bulb = 25 joules

Explanation:

Given:

Energy receive by light bulb = 100 Joules

Energy released by light bulb as light energy = 75 Joules

Find:

Amount of heat energy released by light bulb

Computation:

We know that, energy is neither be created nor destroys

So,

Using Law of conservation of energy

Energy receive by light bulb = Energy released by light bulb as light energy + Amount of heat energy released by light bulb

100 = 75 + Amount of heat energy released by light bulb

Amount of heat energy released by light bulb = 100 - 75

Amount of heat energy released by light bulb = 25 joules

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2 years ago

Balance each of the following redox reactions occurring in basic solution.MnO−4(aq)+Br−(aq)→MnO2(s)+BrO−3(aq)Express your answer

Ahat [919]

Answer : The balanced chemical equation is,

$2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)$

Explanation :

Rules for the balanced chemical equation in basic solution are :

First we have to write into the two half-reactions.
Now balance the main atoms in the reaction.
Now balance the hydrogen and oxygen atoms on both the sides of the reaction.
If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.
If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion $(OH^-)$ at that side where the less number of hydrogen are present.
Now balance the charge.

The half reactions in the basic solution are :

Reduction : $MnO_4^-(aq)+2H_2O(l)+3e^-\rightarrow MnO_2(s)+4OH^-(aq)$ ......(1)

Oxidation : $Br^-(aq)+6OH^-(aq)\rightarrow BrO_3^-(aq)+3H_2O(l)+6e^-$ .......(2)

Now multiply the equation (1) by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)$

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2 years ago

For the following reaction, 28.6 grams of zinc oxide are allowed to react with 9.54 grams of water . zinc oxide(s) water(l) zinc

maw [93]

Answer:

34.9 g of Zn(OH)₂ is the maximum mass that can be formed

Explanation:

Let's state the reaction:

ZnO(s) + H₂O(l) → Zn(OH)₂ (aq)

First of all, we need to determine the moles of each reactant and state the limiting:

28.6 g . 1mol /81.38 g = 0.351 moles of ZnO

9.54 g . 1mol /18 g = 0.53 moles of water

As ratio is 1:1, for 0.53 moles of water, we need 0.53 moles of ZnO, but we only have 0.351, so the limiting reactant is the ZnO.

Ratio with the product is also 1:1. From 0.351 moles of oxide we can produce 0.351 moles of hydroxide. Let's calculate the mass:

0.351 mol . 99.4 g /1mol = 34.9 g

3 0

2 years ago