A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C.What is the specific heat of the substance?

Answers

Answer 1

Answer:

0.3832

Explanation:

Givens

m = 437.2 grams

c = ?

delta t = 69.8 - 19.3

H = 8460 J

Formula

H = m*c*Δt

Solution

8460 = 437.2 * c * (69.8 - 19.3)

8460 = 437.2 * c * 50.5

8460 = 22078.6 * c

c = 8460 / 22078,6

c = .3832 J / (oC * gr)


Related Questions

DUE IN 1 MINUTE HELP I'M DESPERATE

Answers

I believe the answer is A.
The answer will be a

Choose all the answers that apply.
The atmosphere:
is made mostly of nitrogen
can be used to transmit radio signals
traps heat from the sun
protects the earth from harmful radiation
is important in the water cycle

Answers

Answer:1 2 and 4

Explanation:

If a piece of silver specific heat .2165 j/g °C with a mass of 14.16 g and a temperature of 133.5°C is dropped into 250.0 g of fat 17.20°C what will be the final temperature of the system

Answers

Answer:

[tex]T_F=17.56\°C[/tex]

Explanation:

Hello there!

In this case, for this calorimetry problem, it is possible for us to realize that the heat lost by the hot silver is gained by the cold far whose specific heat is 3.94 J/g°c, so we can write:

[tex]-Q_{Ag}=Q_{fat}[/tex]

Which can be written in terms of mass, specific heat and temperature as shown below:

[tex]-m_{Ag}C_{Ag}(T_F-T_{Ag})=m_{fat}C_{fat}(T_F-T_{fat})[/tex]

In such a way, solving for the final temperature, we obtain:

[tex]T_F=\frac{m_{Ag}C_{Ag}T_{Ag}+m_{fat}C_{fat}T_{fat}}{m_{Ag}C_{Ag}+m_{fat}C_{fat}}}[/tex]

Then, we plug in the given data to obtain:

[tex]T_F=\frac{14.16g*0.2165J/g\°C*133.5\°C+250g*3.94J/g\°C*17.20\°C}{14.16g*0.2165J/g\°C+250g*3.94J/g\°C} \\\\T_F=17.56\°C[/tex]

Best regards!

According to the equation below, if 2.00 g of PCl5 react completely, how many grams of HCl will be produced?


PCl5+4H2O→H3PO4+5HCl

Answers

Answer:

1.75 g HCl

Explanation:

PCl₅ + 4H₂O → H₃PO₄ + 5HCl

First we convert 2.00 g of PCl₅ into moles, using its molar mass:

2.00 g ÷ 208.24 g/mol = 0.0096 mol PCl₅

Then we convert PCl₅ moles into HCl moles, using the stoichiometric coefficients of the equation:

0.0096 mol PCl₅ * [tex]\frac{5molHCl}{1molPCl_5}[/tex] = 0.048 mol HCl

Finally we convert HCl moles into grams, using its molar mass:

0.048 mol HCl * 36.45 g/mol = 1.75 g HCl

The precision of a method is being established, and the
following
data are obtained 22.23, 22.18 22.25, 22.09
and 22.15%, is 22.09% a valid measurement?​

Answers

Answer:

No,  22.09%  is not a valid measurement

Explanation:

Precision has to do with how close a given set of measured values are to each other. It is quite different from accuracy. Accuracy refers to how close a given set of values is to the true value. A given set of values may be precise but not accurate and vice versa.

If we look at the values obtained;  22.09%,  22.15%, 22.18%, 22.23%, 22.25%, the value 22.09% is too far off the other values. This implies that it does not represent a valid measurement since it is not close to all the other values obtained.

Pls someone help me with this question pls

Answers

Answer:

So confusing but I'll try

For each of these pairs of half-reactions, write the balanced equation for the overall cell reaction and calculate the standard cell potential. Express the reaction using cell notation. You may wish to refer to Chapter 20 to review writing and balancing redox equations.
1.
Pt2+(aq)+2e-Pt(s)
Sn2+(aq)+2e-Sn(s)
2.
Co2+(aq)+2e-Co(s)
Cr3+(aq)+3e-Cr (s)
3.
Hg2+(aq)+2e-Hg (I)
Cr2+(aq)+2e-Cr (s)

Answers

1. The standard cell potential for this reaction is 0.14 V.

2. The standard cell potential for this reaction is 0.46 V.

3. The reduction potential for Hg2+(aq) + 2e^- → Hg(l) is 0.79 V.

1. The half-reactions are:

Oxidation: Sn2+(aq) → Sn(s) + 2e^-

Reduction: Pt2+(aq) + 2e^- → Pt(s)

To balance the charges, we multiply the oxidation half-reaction by 2:

2Sn2+(aq) → 2Sn(s) + 4e^-

Now, we can combine the half-reactions to form the overall cell reaction:

2Sn2+(aq) + Pt2+(aq) → 2Sn(s) + Pt(s)

The cell notation for this reaction is:

Sn(s) | Sn2+(aq) || Pt2+(aq) | Pt(s)

To calculate the standard cell potential (E°), we can look up the reduction potentials for each half-reaction. The reduction potential for Pt2+(aq) + 2e^- → Pt(s) is typically listed as 0.00 V. The reduction potential for Sn2+(aq) + 2e^- → Sn(s) is -0.14 V. The standard cell potential is the sum of the reduction potentials:

E° = E°(reduction) - E°(oxidation)

E° = 0.00 V - (-0.14 V) = 0.14 V

2. The half-reactions are:

Oxidation: Co2+(aq) → Co(s) + 2e^-

Reduction: Cr3+(aq) + 3e^- → Cr(s)

To balance the charges, we multiply the reduction half-reaction by 2:

2Cr3+(aq) + 6e^- → 2Cr(s)

Now, we can combine the half-reactions to form the overall cell reaction:

Co2+(aq) + 2Cr3+(aq) + 6e^- → Co(s) + 2Cr(s)

The cell notation for this reaction is:

Co(s) | Co2+(aq) || Cr3+(aq) | Cr(s)

To calculate the standard cell potential (E°), we look up the reduction potentials for each half-reaction. The reduction potential for Co2+(aq) + 2e^- → Co(s) is typically listed as -0.28 V. The reduction potential for Cr3+(aq) + 3e^- → Cr(s) is -0.74 V. The standard cell potential is the sum of the reduction potentials:

E° = E°(reduction) - E°(oxidation)

E° = -0.28 V - (-0.74 V) = 0.46 V

3. The half-reactions are:

Oxidation: Cr2+(aq) → Cr(s) + 2e^-

Reduction: Hg2+(aq) + 2e^- → Hg(l)

The balanced overall cell reaction is:

Cr2+(aq) + 2Hg2+(aq) + 4e^- → Cr(s) + 2Hg(l)

The cell notation for this reaction is:

Hg(l) | Hg2+(aq) || Cr2+(aq) | Cr(s)

To calculate the standard cell potential (E°), we look up the reduction potentials for each half-reaction. The reduction potential for Cr2+(aq) + 2e^- → Cr(s) is typically listed as -0.91 V.

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Show the equation you will use to calculate the volume of 1 M Cu(NO3)2 (aq) needed to prepare a set of solutions that have concentrations in the range of 1 M to 1x10-4 M in a 10-mL volumetric flask. Write the reduction half-cell reaction for the copper(II) ion. What is the standard potential for an electrochemical cell that is prepared from a copper half-cell and a zinc half-cell

Answers

Answer:

Explanation:

The equation we use to calculate the volume needed to prepare other [tex](C_1,V_1)[/tex] the solution that has a concentration [tex]C_2[/tex]  and volume [tex]V_2[/tex]  is:

[tex]C_1V_1 =C_2V_2[/tex]

[tex]V_1=\dfrac{C_2V_2}{C_1}[/tex]

where;

[tex]C_1[/tex]= concentration of the first solution

[tex]V_1[/tex] = volume of the first solution

[tex]C_2[/tex] = concentration of the second solution

[tex]V_2[/tex] = volume of the second solution

2) Reduction half cell reaction for the copper (II) ion is:

[tex]Cu^{2+} + 2e^- \to Cu[/tex]

3) [tex]Cu^{+2} + 2e^- \to Cu \text{ \ \ \ E = 0.3370}[/tex]

[tex]Zn^{+2} + 2 e^- \to Zn \ \ \ \ \ \ E = -0.763[/tex]

[tex]Zn \to Zn^{+2} + 2 e^- \ \ \ \ \ \ E = +0.763[/tex]

Since the reduction potential of Cu is more; it means copper will go into reduction and zinc will undergo oxidation.

Standard Potential =[tex]E^0_{left} - E^0_{right}[/tex]

[tex]= -0.763 -0.337[/tex]  ( since both are reduction potential)

[tex]\mathbf{E^0_{cell} = -1.100 volt}[/tex]

From these four cycles which are water cycle, carbon cycle, nitrogen cycle, and phosphorus cycle which cycle has more nutrients

Answers

Answer:

Nitrogen cycle

Explanation:

Someone help thank you :)

Answers

The link to this question is www.com

How many molecules of N204 are in 85.0 g of N2O4?

Answers

Answer:

5.56 x 10^23

Explanation:

Just convert and cancel out.

85 g N2O4 x 1 mol/92.01 g x 6.02 x 10^23 molecules /1 mol

In an experiment, 135.2 grams of a mystery metal is heated to 100.0*C. The metal is then dumped into a calorimeter with 59.0 grams of water at 26.0*C. The temperature of the water in the calorimeter increases to 35.0*C after the metal is dumped into it. What is the specific heat capacity of the metal? Express your answer to 2 past the decimal. Do NOT include units

Answers

The answer wound be 23 units I think

Write a short essay about life in the Han Dynasty, comparing it to life today. Make sure to include key features:
-Family
-Government
-Social Structure
-Religion
-Trade

Answers

Answer:

Life in the Han Dynasty (206 BCE - 220 CE) differed significantly from today in family, government, social structure, religion, and trade. For example, the Han Dynasty emphasized a patriarchal family structure, where the eldest male held authority, and filial piety was highly valued. In contrast, contemporary societies embrace more egalitarian family dynamics with shared decision-making.

The government system of the Han Dynasty relied on a centralized bureaucracy and emphasized meritocracy, while modern societies often adopted democratic systems. Socially, the Han Dynasty followed a hierarchical model influenced by Confucian principles, whereas contemporary societies strive for greater equality and social mobility.

Religion in the Han Dynasty combined Confucianism, Taoism, and Buddhism, whereas modern societies exhibit diverse religious beliefs. Lastly, trade in the Han Dynasty thrived along the Silk Road, while modern trade was globally interconnected and facilitated by technological advancements. These differences highlight the evolution of society over time.

Explanation:

Plants are divided into three groups based on the

Answers

Answer:Scientists have identified more than 260,000 kinds of plants. They classify plants according to whether they have body parts such as seeds, tubes, roots, stems, and leaves. The three main groups of plants are seed plants, ferns, and mosses.

Explanation:

hope i help

A 10.0 mL sample of HNO3 was diluted to a
volume of 100.00 mL. Then 25 mL of that
diluted solution was needed to neutralize 50.0
mL of 0.60 M KOH. What was the
concentration of the original nitric acid?
1.2 M
12 M
none of these
O 0.12 M
0.0012M

Answers

Answer:

12 M

Explanation:

The reaction between HNO₃ and KOH is:

HNO₃ + KOH → KNO₃ + H₂O

First we calculate how many KOH moles reacted with the diluted HNO₃ sample, using the given volume and concentration:

50.0 mL * 0.60 M = 30 mmol KOH

As 1 KOH mol reacts with 1 HNO₃ mol, in 25 mL of the diluted HNO solution there are 30 HNO mmoles.

With that information in mind we can calculate the HNO₃ concentration in the diluted solution:

30 mmol HNO₃ / 25 mL = 1.2 M

Finally we can use the C₁V₁=C₂V₂ formula to calculate the concentration of the original solution:

C₁ * 10.0 mL = 1.2 M * 100.00 mLC₁ = 12 M

How do I solve stoichiometry problems? Please I need help with at least one of these stoichiometry problems with work and steps so I understand better. Thank you!

Answers

Answer:

2. m = 426.6 gr

4. m = 143 gr

What kind of waste is shown in the photograph?
A. Composted waste
B. E-waste
C. Hazardous waste
O D. Septic system waste

Answers

it should be C. hazardous waste
hope this helps!!

a Technical
How to write
report​

Answers

Excuse me? ????????????????

Use the following Balanced Equation to complete the question: 2 Al + 6 HBr → 2 AlBr3 + 3 H2

If you have 10 moles of Al how many moles of H2 can be produced?


*Will give Brainly!*

Answers

Here is the answer enjoy

Many bones get their name from the bone they are

Answers

What? What do you mean here?

List three ways in which the octet rule can sometimes fail to be obeyed

Answers

Answer:

1. Electron-deficient molecules

2. Odd electron molecules

3. Expanded valence shell molecules

Explanation:

The trait that seems to fade into the background.
Dominant
Recessive

Answers

The correct answer is Recessive
Recessive I guess? I don’t even know what this mean

Tarnish on silver is the compound Ag2OAg2O. A tarnished silver spoon is placed in an aluminum pan of boiling water. When enough salt is added so that the solution conducts electricity, the tarnish disappears. Imagine that the two halves of this redox reaction were separated and connected with a wire and a salt bridge.

Answers

Answer:

The answer is "2.459".

Explanation:

The cell potential or emf norm  = oxidation pot. of anode -oxidation pot. of cathode

When the pot oxidizer is the opposite of a red pot size it's been transformed to either the redox reaction so that anode was its higher oxide pot material that means (AL) so,

[tex]\to emf=1.66-(-0.799) \\\\ \to emf=1.66+0.799) \\\\ \to emf =2.459[/tex]

How many moles of NaOH are contained in 56.0 mL of a 2.40 M solution of 1 point
NaOH in water? (**Use only numerical answers with 3 significant figures.
The units are given in the question.)
Your answer

Answers

Answer:

1.34 mol

Explanation:

Molarity, which is the molar concentration of a solution, can be calculated by dividing the number of moles (n) by the volume (V).

That is;

Molarity (M) = n/V

According to the information provided in this question;

M = 2.40M

V = 56.0 mL = 56/1000 = 0.056 L

Since molarity = n/V

number of moles = M × V

n = 0.056 × 24

n = 1.34 mol

Why is SpaceX likely to succeed in a mission to Mars?
A. It only hires expert NASA employees as its employees.
B. It takes more risks than NASA, which cannot afford them.
C. It is run by Elon Musk, who is determined to get to Mars.
D. It has fewer restrictions than NASA does.

Answers

Answer:

C

Explanation:

It is run by Elon Musk, who is determined to get to Mars.

Suppose that you have 70 grams of calcium chloride (CaCl2) contaminated with 5% sodium chloride (NaCl) and 2% sand. You dissolve the impure solid in 54 ml of hot water, and allow it to cool and crystallize. What would be the maximum mass (grams) of purified CaCl2 that would you would be able to recover

Answers

Answer:

65.1 grams

Explanation:

The percentage of impurities in the contaminated CaCl₂ sample is:

% of NaCl + % of sand = 7%

Meaning that the purity of the sample is (100 - 7) 93%.

Now we calculate the mass of pure CaCl₂ in the contaminated sample:

70 g * 93/100 = 65.1 g

Thus the answer is 65.1 grams, as that amount is how much CaCl₂ is in the sample and it would be impossible to obtain more than that.


20 points!

Which two structures produce energy that cells can
use?

A and B

B and C

C and D


D and A

Answers

Answer:

c and d

Explanation:

mitochondria and vacoule

Answer:

It is actually D and A

Explanation:

I took this assignment before on edge and the guy who said C and D is wrong

Do 5. What is the total number of atoms contained in a 1.00-mole sample of
helium?

Answers

Answer: 1 mole =  6.02 x 10^ 23

Explanation:

The number of atoms in one mole of helium is equal to 6.022×10²³ He atoms.

What is Avogadro's number?

Avogadro’s constant can be demonstrated as the proportionality factor that utilizes to count the number of particles such as ions, molecules, atoms, or ions in a sample of the substance.

Avogadro's number can be defined as the approximate count of nucleons in 1 gram of substance. The value of the Avogadro number is the mass of 1 mole of a compound, in grams, and is the number of nucleons in one particle.

The value of Avogadro’s number is equal to 6.022×10²³ per mole.

Given, the number of moles of the helium = 1 mol

Helium is a monatomic gas so it contains only one atom in its one molecule.

The number of atoms of Helium in one mole = 6.022 × 10²³

Therefore, the total number of atoms contained in a 1 mole sample of

helium is 6.022 × 10²³.

Learn more about Avogadro's number, here:

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What would you do during a zombie apocalypse
A. run and hide
B. fight back
C. save some people
D. raid survivor's homes
E. keep your family alive

Answers

Answer:

C

Explanation:

because then you'd be able to make a group to raid, run with you, and save more people.

A moving 10 kilogram object has 20 Joules of kinetic energy. What is its velocity?
4 m/s
2 m/s
10 m/s
200 m/s

Answers

It’s the second one you’re welcome
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