Determine the formula for iron l oxide

A scientist is studying the effect of climate change on a population of fur seals living on an island off the coast of Alaska. F

solniwko [45]

Answer:

A

Explanation:

The only factor that would not decrease the carrying capacity of the fur seal's environment is the population of predators.

The carrying capacity is defined as the maximum number of individuals of a species an environment can sustain based on the resources available. Hence, only a fluctuation in the resources available in the environment of the fur seal can affect the carrying capacity.

Pollution and destruction of the habitat of the fur seal would limit the resources available to the animals and as such, limit the carrying capacity. A fluctuation in the population of prey available for the population of fur seals would also affect the carrying capacity.

The only factor that has no bearing on the carrying capacity of the fur seal's environment is the population of predators. Predators have the capacity to reduce the population of fur seals but not affect the carrying capacity for fur seals.

The correct option is, therefore, A.

3 0

3 years ago

10.0 grams of octane are burned in a bomb calorimeter containing 2.00 × 102 g H2O. How much energy, in kilocalories, is released

vodomira [7]

you wanna use q=m*C*\Delta T\;

where Delta T=T_f-T_i for m, you wanna use the mass of the water not the octane because you're measuring the temp change of the water.

6 0

4 years ago

A compound of formula XCl3 reacts with aqueous AgNO3 to yield solid AgCl according to the following equation: XCl3(aq)+3AgNO3(aq

Levart [38]

Answer:

Aluminum

Explanation:

$Moles =\frac {Given\ mass}{Molar\ mass}$

Mass of $AgCl$ = 1.68 g

Molar mass of $AgCl$ = 143.32 g/mol

Thus,

$Moles\ of\ AgCl=\frac {1.68}{169.87}=0.01172$

From the reaction below:

$XCl_3_{(aq)}+3AgNO_3_{(aq)}\rightarrow X(NO_3)_3_{(aq)}+3AgCl_{(s)}$

3 moles of $AgCl$ are produced when 1 mole of $XCl_3$ undergoes reaction.

So,

1 mole of $AgCl$ are produced when $\frac {1}{3}$ mole of $XCl_3$ undergoes reaction.

0.01172 mole of $AgCl$ are produced when $\frac {1}{3}\times 0.01172$ mole of $XCl_3$ undergoes reaction.

Thus, moles of $XCl_3$ = 0.0039 moles

Let the atomic mass of X = x g/mol

atomic mass of chlorine = 35.5 g/mol

Thus, Molar mass of $XCl_3$ = x + 3(35.5) g/mol = x + 106.5 g/mol

Moles = 0.0039 moles

Mass = 0.521 g

Thus, molar mass = Given mass/ Moles = 0.521 / 0.0039 = 133.5897 g/mol

So,

x + 106.5 = 133.5897

x = 27.0897 g/mol

This Atomic weight corresponds to Aluminum. Hence, X is aluminum.

8 0

4 years ago

Can someone help me with this please :)

antoniya [11.8K]

Answer: Hydrochloric acid poses a various health risk

5 0

3 years ago

A 8.06 g piece of solid CO 2 is allowed to sublime in a balloon. The final volume of the balloon is 1.00 L at 300.0 K. What is t

natka813 [3]

Answer:

The pressure of the gas is 4.428 atm.

Explanation:

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

Considering a certain amount of ideal gas confined in a container where the pressure, volume and temperature can vary, but keeping the mass constant, that is, without altering the number of moles, the pressure, P, the temperature can be related, T and the volume, V, of an ideal gas using the ideal gas law:

P*V = n*R*T

where R is the ideal gas constant, and n is the number of moles of the gas.

In this case, to know n you must know the molar mass of the CO₂ compound. Being:

C: 12 g/mole
O: 16 g/mole

then, the molar mass of CO₂ is: 12 g/mole + 2*16 g/mole= 44 g/mole

Then you apply a rule of three: if 44 grams of CO₂ are present in 1 mole, 8.06 grams in how many moles are they?

$moles of CO_{2} =\frac{8.06 grams*1 mole}{44 grams}$

moles of CO₂= 0.18 moles

Then, you know:

P: ?
V: 1 L
n: 0.18 moles

R= 0.082 $\frac{atm*L}{mol*K}$

T= 300 K

Replacing in the equation of the ideal gas law:

P* 1 L= 0.18 moles* 0.082 $\frac{atm*L}{mol*K}$ * 300 K

Solving:

$P=\frac{0.18 moles* 0.082 \frac{atm*L}{mol*K} *300 K}{1 L}$

P= 4.428 atm

The pressure of the gas is 4.428 atm.

3 0

3 years ago