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4 years ago

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Chemistry

1 answer:

matrenka [14]4 years ago

4 0

Answer:

A. tissues are comprised of cells

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In an oxidation-reduction reaction, the reducing agent In an oxidation-reduction reaction, the reducing agent gains electrons an

Leto [7]

Answer:

loses electrons and loses potential energy.

Explanation:

A reducing agent looses electrons in an oxidation-reduction reaction. Oxidation is defined as the loss of electrons.

Reducing agents are oxidized in an oxidation-reduction reaction. When a specie looses electrons, it also looses energy. Hence the answer given above.

8 0

3 years ago

Your calculated density of aluminum is d = 2.69 g/cm3. Aluminum’s accepted density is 2.70 g/cm3. Without writing the "%" sign,

stepan [7]

Answer:

0.37 %

Explanation:

Given that:

Calculated density of aluminum = 2.69 g/cm³

Accepted density of aluminum = 2.70 g/cm³

$Error\ percentage=\frac {|Accepted\ value-Calculated\ value|}{Accepted\ value}\times 100$

Thus, applying values as:

$Error\ percentage=\frac {|2.70-2.69|}{2.70}\times 100$

<u>Percent error = 0.37 %</u>

7 0

3 years ago

Read 2 more answers

The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C

artcher [175]

Explanation:

The given data is as follows.

$P_{1}$ = 100 mm Hg or $\frac{100}{760}atm$ = 0.13157 atm

$T_{1}$ = $1080 ^{o}C$ = (1080 + 273) K = 1357 K

$T_{2}$ = $1220 ^{o}C$ = (1220 + 273) K = 1493 K

$P_{2}$ = 600 mm Hg or $\frac{600}{760}atm$ = 0.7895 atm

R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

$log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}$

$log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]$

$log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]$

0.77815 = $\frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K$

$\Delta H_{vap}$ = $2.219 \times 10^{5}$ J/mol

= $2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}$

= 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

4 0

3 years ago

What is the density of a rectangle solid with dimensions

Hunter-Best [27]

Answer:

5625000gm/cm^3

Explanation:

volume=w*h*l=25*15*5=1875cm^3

density=mass*volume

=1875*3000

=5625000gm/cm^3

8 0

3 years ago

How many atoms of hydrogen are in 100 g of hydrogen peroxide (h2o2)?

Maurinko [17]

Hello friends..

find the MW of HP
calculate the # of mols per 100g of HP
take the # of mols times 6.023 x 10^23 times 2 (2 H per molecule)

Hope it helps you..

7 0

3 years ago