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2 years ago

7. Fredrick wanted to find out what soil works best for growing roses. He grew them in potting

Chemistry

1 answer:

Lady_Fox [76]2 years ago

5 0

Answer:

Dependent variable are those variables that changes with the change in variables that affects it. (IG - leo_muan)

Explanation:

Dependent variable responds or gets affected to change. They are not constant. It can be possible that dependent variable is relying on independent variable as independent variable does not change with change of any of the factor in an experiment.

Sometimes, mentioning not the units leads to insufficient data of the variables. A constant is something whose value cannot be changed and is used for comparison purpose. According to the question, Rose is constant, height is dependent variable , different types of soil are independent variable.

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A sample of 4.0 L of nitrogen, at 1.2 atmospheres, is transferred to a 12 L container.

Zielflug [23.3K]

Answer:

The pressure will be 0.4 atm.

Explanation:

The gas laws are a set of chemical and physical laws that allow determining the behavior of gases in a closed system. The parameters evaluated in these laws are pressure, volume, temperature and moles.

As the volume increases, the gas particles (atoms or molecules) take longer to reach the walls of the container and therefore collide with them less times per unit of time. This means that the pressure will be lower because it represents the frequency of collisions of the gas against the walls. In this way pressure and volume are related, determining Boyle's law which says:

"The volume occupied by a certain gaseous mass at constant temperature is inversely proportional to pressure"

Boyle's law is expressed mathematically as:

P*V= k

If you initially have the gas at a volume V1 and press P1, when the conditions change to a volume V2 and pressure P2, the following is satisfied:

P1*V1= P2*V2

In this case:

P1= 1.2 atm
V1= 4 L
P2= ?
V2= 12 L

Replacing:

1.2 atm* 4 L= P2* 12 L

Solving:

$P2=\frac{1.2 atm*4 L}{12 L}$

P2= 0.4 atm

<u><em>The pressure will be 0.4 atm.</em></u>

7 0

2 years ago

If 20 mL of gas is subjected to a temperature change from 10°C to 100°C and a pressure change from 1 atm to 10 atm,

Evgesh-ka [11]

Answer:

E. None of these

Explanation:

We know, By GAS laws,

PV = NRT, where p- pressure, v- volume, n- number of moles, R- gas constant ,and T- temperature

Now, In the question, the number of moles remains the same as the gas is the same. so n is constant so we can compare n before and after a temperature change.

$\frac{P1V1}{RT1}$ = $\frac{P2V2}{RT2}$

where P1= 1 atm, P2 = 10 atm, V1= 20 mL, T1= 10°C and T2= 100°C

We don't have to worry about the standard units as they are present equally on both the sides and get cut, same goes for R( gas constant)

So putting values, we get

$\frac{1*20}{R*10} = \frac{10*V2}{R*100}$

Cutting, R on both sides and moving contents to the right so that only V2 is left on the left.

$\frac{1*20*100}{10*10} = V2$

∴ V2 = $\frac{2000}{100}$

∴ V2 = 20mL

3 0

1 year ago

Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C

ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)

The enthalpy of reaction (1) is given by:

$\Delta H = \Delta H_{r} - \Delta H_{p}$ (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond

7O₂: 7 moles of 1 O=O bond

4CO₂: 4 moles of 2 C=O bonds

6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

$\Delta H = \Delta H_{r} - \Delta H_{p}$

$\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})$

$\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})$

$\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol$

$\Delta H = -2860 kJ/mol$

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

I hope it helps you!

7 0

2 years ago

Which is the correct name for p2o5? phosphorus dioxide phosphorus pentoxide diphosphorus pentoxide diphosphorus hexaoxide

Elenna [48]

Correct option:

The correct name for $P_{2}O_{5}$ is diphosphorus pentoxide.

Why $P_{2}O_{5}$ is called diphosphorus pentoxide?

$P_{2}O_{5}$ is commonly known as diphosphorus pentoxide.

Phosphorus pentoxide has an intriguing property in that $P_{2}O_{5}$ is actually its empirical formula, whereas $P_{4}O_{10}$ is its actual molecular formula.

However, the name of the chemical was obtained from its empirical formula rather than from its molecular formula. The official name for this substance is diphosphorus pentoxide.

Oxygen-containing binary compounds have "oxide" as their "last name." Phosphorus is the "first name."

We list each atom's numbers below:

The di- and Penta- prefixes are used to indicate the presence of two and five oxygen atoms, respectively, in the molecule.

Learn more about diphosphorus pentoxide here,

brainly.com/question/18237346

#SPJ4

6 0

2 years ago

How many liters of CO2 are formed when 14.0 g of CaCO3 react at 1.00 atm and 1000K

Gnesinka [82]

Answer: 11.5

Explanation: