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2 years ago

A red hot iron bar is dipped into 600 ml of water ( 600.g of water) which causes

Chemistry

1 answer:

Novosadov [1.4K]2 years ago

4 0

Answer:

168277.2 J

Explanation:

m = Mass of water = 600 g

c = Specific heat of water = 4.186 J/g°C

$\Delta T$ = Change in temperature of water = $89-22=67^{\circ}\text{C}$

We know that the amount of heat lost = amount of heat gained

$Q=mc\Delta T\\\Rightarrow Q=600\times 4.186\times 67\\\Rightarrow Q=168277.2\ \text{J}$

Heat lost by the iron bar is 168277.2 J.

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Explain how the following changes would affect the rate of the reaction of 2-bromo-3-methylbutane with methanol: Part A The alky

swat32

Answer:

The reaction will be slower because the leaving group will be poorer.

Explanation:

One Important thing to put at the back of our mind is that weak bases are good leaving groups. Another thing to take note is that in the halogen series(which is our main subject in this question) as you go down the group the greater or the more heavy the halide is and the heavier halides are more stable that is Bromine will be more stable than chlorine.

Now, we are now told that the bromo halide is been replaced by the chloro halide which means that the rate of Reaction will surely decrease because the leaving group. The cl^- is a stronger base.

6 0

2 years ago

The mass of a block is 12 grams and the volume is 6 ml. Calculate the density of the

KengaRu [80]

Answer:

The density of block is 2 g/mL.

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Symbol:

The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.

Given data:

mass of block = 12 g

volume = 6 mL

density = ?

Now we will put the values in the formula,

d= m/v

d = 12 g/ 6 mL

d = 2 g/mL

so, the density of block is 2 g/mL .

3 0

2 years ago

How many moles are in 7.2x10^15 atoms of Pb?

fiasKO [112]

<h3>Answer:</h3>

1.2 × 10⁻⁸ mol Pb

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 7.2 × 10¹⁵ atoms Pb

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 7.2 \cdot 10^{15} \ atoms \ Pb(\frac{1 \ mol \ Pb}{6.022 \cdot 10^{23} \ atoms \ Pb})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 1.19562 \cdot 10^{-8} \ mol \ Pb$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.19562 × 10⁻⁸ mol Pb ≈ 1.2 × 10⁻⁸ mol Pb

6 0

2 years ago

The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal

spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0

2 years ago

CONCLUSIONS: Suppose a student titrated a sample of monoprotic acid of unknown concentration using a previously standardized sol

Zolol [24]

Concentration of unknown acid is 0.061 M

Given:

Concentration of NaOH = 0.125 M

Volume of NaOH = 24.68 mL

Volume of acid solution = 50.00 mL

To Find:

Concentration of the unknown acid

Solution: Concentration is the abundance of a constituent divided by the total volume of a mixture. The concentration of the solution tells you how much solute has been dissolved in the solvent

Here we will use the formula for concentration:

M1V1 = M2V2

0.125 x 24.68 = 50 x M2

M2 = 0.125 x 24.68 / 50

M2 = 0.061 M

Hence, the concentration of unknown acid is 0.061 M

Learn more about Concentration here:

brainly.com/question/17206790

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3 0

1 year ago