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4 years ago

A 350. mL air sample collected at 35 degrees C has a pressure of 556 mm Hg.

Chemistry

1 answer:

SashulF [63]4 years ago

7 0

Answer:

485 torr

Explanation:

Explanation:

We can solve this equation using the combined gas law:

P1V1/T1=P2V2/T2

Remember, always use the absolute (Kelvin) temperature when working with gas equations.

The temperature conversions are

T1=C+273=35o+273=308K

T2=57C+273=330K

Since we're trying to find the final pressure, let's rearrange this equation to solve for

P2:

P2=P1V1T2/T1V2

Our known values:

P1= 550 torr

V1=350mL

T1=308K

V2=425"mL

T2=330K

Let's plug these into the equation to find the final pressure:

The final pressure after it is subjected to these changes is thus 485torr

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Characteristic blue eyes and blond hair?

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Monosodium glutamate (NaC5H8NO4) is used extensively as a flavor enhancer what is its percent composition

tensa zangetsu [6.8K]

Answer:

% sodium= 13.6 % sodium

% carbon= 35.5 % carbon

% hydrogen= 4.7% hydrogen

% nitrogen = 8.3% nitrogen

% oxygen = 37.8 % oxygen

Explanation:

To find its percent composition means that we are to find to find the percentage of each of the constituents of the compound present.

The molar mass of monosodium glutamate is 169.11 gmol-1

Hence;

Percent of sodium= 23 gmol-1/169.11 gmol-1 × 100 = 13.6 % sodium

Percent of Carbon= 60 gmol-1/169.11 gmol-1 ×100 = 35.5 % carbon

Percent of hydrogen= 8/169.11 gmol-1 ×100 = 4.7% hydrogen

Percent nitrogen = 14/169.11 gmol-1 × 100 = 8.3% nitrogen

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3 years ago

Explain why photosynthesis and respiration are considered opposite processes when it comes to carbon.

lyudmila [28]

Respiration and Photosynthesis is different form each other because photosynthesis removes CO2 while respiration puts back CO2

I hoped this helped!

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3 years ago

20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What's the new volu

belka [17]

10 cubic inches
We will use Boyle's law that states that for a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional.
P1 V1 = P2 V2
Where
P1 is initial pressure = 5 psi
V1 is initial volume = 20 cubic inch
P2 is final pressure = 10 psi
V2 is final volume = unknown
V2 = P1,V1 / P2
V2 = 20 × 5 / 10
V2 = 100/10
V2 = 10 cubic inches

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3 years ago

How much energy must be added to a 1 kg piece of gold with a specific heat of 130 j/kg to increase its temperature from 20 c to

Lorico [155]

$Q = 10.4 \; \text{kJ}$

The specific heat of a material gives the energy it takes to increase the temperature of one unit mass of the material by one unit temperature. The SI unit for specific heat $c$ is therefore $\text{J} /\text{kg} \cdot \textdegree{\text{C}}$ .

The formula relating the energy required to raise the temperature of $m$ grams of a substance with specific energy $c$ by $\Delta T$ degrees is

$Q= c \cdot m \cdot \Delta T$

The question provides the following information:

$c = 130\; \text{J} /\text{kg} \cdot \textdegree{\text{C}}$ for gold
$m= 1 \; \text{kg}$
$\Delta T = 100 \textdegree{\text{C}} - 20 \textdegree{\text{C}} = 80 \textdegree{\text{C}}$

Apply the formula:

$\begin{array}{ccc} Q & = & c \cdot m \cdot \Delta T\\ & = & 130 \; \text{J} \cdot \text{kg}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1 \; \text{kg} \times 80 \; \textdegree{\text{C}}\\ & = & 1.04 \times 10^{4} \;\text{J}\\ & = & 10.4 \; \text {kJ} \end{array}$

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4 years ago