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Elena-2011 [213]
3 years ago
9

Calculate the amount of heat released when 27.0 g H2O is cooled from a liquid at 314 K to a solid at 263 K. The melting point of

H2O is 273 K. Other useful information about water is listed below.
Cp,liquid = 4.184 J/(g•K)

Cp,solid = 2.093 J/(g•K)

ΔHfusion = 40.7 kJ/mol
Chemistry
2 answers:
Liula [17]3 years ago
7 0

Answer:

Cool liquid from 314 K to 273 K, freeze liquid at 273 K, and cool solid to 263 K.

Explanation:

answer on edg

lubasha [3.4K]3 years ago
5 0
Cool liquid from 314 K to 273 K, freeze liquid at 273 K, and cool solid to 263 K.
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Calculate the percentage by mass of the indicated element in the following compounds. Hydrogen in ascorbic acid, HC6H7O6, also k
Scorpion4ik [409]

Answer: The mass percent of hydrogen in ascorbic acid is 4.5 %

Explanation:

In C_6H_8O_6, there are 6 carbon atoms, 8 hydrogen atoms and 6 oxygen atoms.

To calculate the mass percent of element in a given compound, we use the formula:

\text{Mass percent of hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Molar mass of ascorbic acid}}\times 100

Mass of hydrogen = 8\times 1g/mol=8g

Molar Mass of ascorbic acid =6\times 12g/mol+8\times 1g/mol+6\times 16g/mol=176g

Putting values in above equation, we get:

\text{Mass percent of hydrogen}=\frac{8g}{176}\times 100=4.5\%

Hence, the mass percent of of hydrogen in ascorbic acid is 4.5 %.

8 0
3 years ago
Can anybody check my answer?
anzhelika [568]

Answer:

\boxed{\text{25. 20 L; 26. 49 K}}

Explanation:

25. Boyle's Law

The temperature and amount of gas are constant, so we can use Boyle’s Law.

p_{1}V_{1} = p_{2}V_{2}

Data:

\begin{array}{rcrrcl}p_{1}& =& \text{100 kPa}\qquad & V_{1} &= & \text{10.00 L} \\p_{2}& =& \text{50 kPa}\qquad & V_{2} &= & ?\\\end{array}

Calculations:

\begin{array}{rcl}100 \times 10.00 & =& 50V_{2}\\1000 & = & 50V_{2}\\V_{2} & = &\textbf{20 L}\\\end{array}\\\text{The new volume will be } \boxed{\textbf{20 L}}

26. Ideal Gas Law

We have p, V and n, so we can use the Ideal Gas Law to calculate the volume.

pV = nRT

Data:  

p = 101.3 kPa

V = 20 L

n = 5 mol

R = 8.314 kPa·L·K⁻¹mol⁻¹

Calculation:

101.3 × 20 = 5 ×  8.314 × T

2026 = 41.57T

T = \dfrac{2026}{41.57} = \textbf{49 K}\\\\\text{The Kelvin temperature is }\boxed{\textbf{49 K}}

6 0
3 years ago
The total volume required to reach the endpoint of a titration required more than the 50 mL total volume of the buret. An initia
Margaret [11]

Answer:

The endpoint volume is 50.52 ±  0.14 mL

Explanation:

In a titration always is necessary to subtract the blank volume to the titrant volume to obtain the real volume of the titrant. Thus in this case, the total endpoint volume is the sum of the initial volume delivered and the second volume delivered, minus the blank volume:

V = (49.16±0.06 mL) + (1.69±0.04 mL) - (0.33±0.04 mL)

V = (49.16 + 1.69 - 0.33) ± (0.06+0.04+0.04) mL

V = 50.52 ±  0.14 mL

It is necessary to consider the sum of the errors too.

7 0
3 years ago
How can we tell the distance sediments traveled due to erosion ?​
meriva

Answer:

more rounded the grains are the more they have been moved around

Explanation:

Generally – the more rounded the grains are the more they have been moved around (i.e. the longer the length of time or distance they have moved). Angular grains cannot have travelled far

geolsoc.org.uk

4 0
2 years ago
How to do q solution, qrxn, moles of Mg , and delta Hrxn?
Helga [31]

Answer:

<em> 14, 508J/K</em>

ΔHrxn =q/n

where q = heat absorbed and n = moles

Explanation:

<em>m = mass of substance (g) = 0.1184g</em>

1 mole of Mg - 24g

<em>n</em> moles - 0.1184g

<em>n = 0.0049 moles.</em>

Also, q = m × c × ΔT

<em> Heat Capacity, C of MgCl2 = 71.09 J/(mol K)</em>

<em>∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)</em>

<em>= 14, 508 J/K/kg</em>

ΔT=  (final - initial) temp = 38.3 - 27.2

= 11.1 °C.

mass of MgCl2 = 95.211 × 0.1184 = 11.27

⇒ q = 11.27g × 11.1 °C × <em>14, 508 j/K/kg </em>

<em>= 1,7117.7472 J °C-1 g-1</em>

<em />

<em>∴ ΔHrxn = q/n</em>

<em>=1,7117.7472  ÷ 0.1184 </em>

<em>= 14, 508J/K</em>

6 0
3 years ago
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