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4 years ago

8

which of the following would be a testable hypothesis for an experimet that compares the health of salmon raisedin fis farms to

salmon caught in the wild?

1 answer:

Alex73 [517]4 years ago

7 0

Which habitat has salmon with less disease per 100 salmon?

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Consider the unbalanced equation for the oxidation of butene. C4H8 + 6O2 Right arrow. CO2 + H2O For each molecule of C4H8 that r

antiseptic1488 [7]

Answer is C4H8 + 6O2 —> 4CO2 + 4H2O

4 0

3 years ago

All atoms of the same element have the same:

Korvikt [17]

They have the same number of protons

8 0

3 years ago

Read 2 more answers

A sample of a gas in a rigid cylinder with a movable piston has a volume of a lamp then .2 L at STP. What is the volume of this

Gnoma [55]

<u>Answer:</u> The new volume of the gas is 0.11 L

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

At STP:

The temperature at this condition is taken as 273 K and the pressure at this condition is taken as 1 atm or 101.3 kPa.

We are given:

$P_1=101.3kPa\\V_1=0.2L\\T_1=273K\\P_2=202.6kPa\\V_2=?\\T_2=300K$

Putting values in above equation, we get:

$\frac{101.3kPa\times 0.2L}{273K}=\frac{202.6kPa\times V_2}{300K}\\\\V_2=\frac{101.3\times 0.2\times 300}{273\times 202.6}=0.11L$

Hence, the new volume of the gas is 0.11 L

3 0

3 years ago

Sharpening a pencil leaves behind pencil shavings. is sharpening a physical or chemical change

IgorLugansk [536]

Physical damage.

If you burned the pencil, it would be considered chemical.

6 0

3 years ago

Read 2 more answers

You start with an equimolar mixture of the gases SO₂ and O₂ in a container. The SO₂ and O₂ react to completion to form SO₃. If t

LuckyWell [14K]

The density of the product gas mixture is 5.39 g/L

Explanation:

<u>Given:</u>

Temperature after the reaction = $25^0$ C

Pressure in the container = 1.65 atm

The ideal gas equation PV = nRT

<u>To find:</u> Density of the product gas mixture

<u>Step 1:</u>

Molarity (M) = P/(RT)

$\[M=\frac{P}{R T}\]$M=\frac{1.65 \mathrm{atm}}{\left(0.08206 \frac{L \cdot a t m}{m o l . K}\right)(298.15 K)}$$

M = 0.0674 mol/L

<u>Step 2: </u>

The product of the given reaction is $SO_3$ . Its molar mass is

1S x 32.066 g/mol = 32.066 g/mol

3O x 16 g/mol = 48 g/mol

Adding both we get 80.066 g/mol

Therefore the density of the product gas mixture is $0.0674 \frac{{mol}}{L} \times \frac{80.066 g}{1 \mathrm{mol}}$

= 5.39 g/L

8 0

3 years ago