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3 years ago

How do conditions change as the depth of the ocean water increases?

Chemistry

2 answers:

Doss [256]3 years ago

7 0

Answer:

Explanation:

natta225 [31]3 years ago

5 0

Answer:

Temperature decreases and pressure increases

Explanation:

The water molecules pack together tighter as pressure increases –the pressure increase with depth, due to the weight of the water above, and causes the greatest density changes in seawater with depth (greater than the density changes due to temperature and salinity changes).

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Iron oxide reacts with aluminum to give aluminum oxide and iron. What kind of chemical reaction is this?

Dima020 [189]

FeO + Al -> AlO + Fe

As you can see, this type of reaction would lead to be a Single Placement (or Replacement) reaction!

Hope this helps! Best wishes!

3 0

3 years ago

I need help on both a and b of question 1

marishachu [46]

Answer:

(a) -0.00017 M/s;

(b) 0.00034 M/s

Explanation:

(a) Rate of a reaction is defined as change in molarity in a unit time, that is:

$r = \frac{\Delta c}{\Delta t}$

Given the following reaction:

$2 N_2O_5 (g)\rightleftharpoons 4 NO_2 (g) + O_2 (g)$

We may write the rate expression in terms of reactants firstly. Since reactants are decreasing in molarity, we're adding a negative sign. Similarly, if we wish to look at the overall reaction rate, we need to divide by stoichiometric coefficients:

$r = -\frac{\Delta [N_2O_5]}{2 \Delta t}$

Reaction rate is also equal to the rate of formation of products divided by their coefficients:

$r = \frac{\Delta [NO_2]}{4 \Delta t} = \frac{\Delta [O_2]}{\Delta t}$

Let's find the rate of disappearance of the reactant firstly. This would be found dividing the change in molarity by the change in time:

$r_{N_2O_5} = \frac{0.066 M - 0.100 M}{200.00 s - 0.00 s} = -0.00017 M/s$

(b) Using the relationship derived previously, we know that:

$-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}$

Rate of appearance of nitrogen dioxide is given by:

$r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t}$

Which is obtained from the equation:

$-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}$

If we multiply both sides by 4, that is:

$-\frac{4 \Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{\Delta t}$

This yields:

[tex]r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t} = -2\frac{\Delta [N_2O_5]}{ \Delta t} = -2\cdot (-0.00017 M/s) = 0.00034 M/s[tex]

5 0

3 years ago

The two common chlorides of phosphorus, PCl3, and PCl5, both important for the production of the other phosphorus compounds, coe

Irina18 [472]

Answer:

For a: The value of $K_c$ for the given reaction is 271.6

For b: The value of $K_p$ for the reaction is 6.32

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $PCl_5$ :

Given mass of $PCl_5$ = 0.105 g

Molar mass of $PCl_5$ = 208.24 g/mol

Putting values in equation 1, we get:

$\text{Moles of }PCl_5=\frac{0.105g}{208.24g/mol}=5.04\times 10^{-4}mol$

For $PCl_3$ :

Given mass of $PCl_3$ = 0.220 g

Molar mass of $PCl_5$ = 137.33 g/mol

Putting values in equation 1, we get:

$\text{Moles of }PCl_3=\frac{0.220g}{137.33g/mol}=1.60\times 10^{-3}mol$

For $Cl_2$ :

Given mass of $Cl_2$ = 2.12 g

Molar mass of $Cl_2$ = 71.0 g/mol

Putting values in equation 1, we get:

$\text{Moles of }Cl_2=\frac{2.12g}{71.0g/mol}=0.029mol$

Volume of the flask = 25.0 L

For the given chemical equation:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

For a:

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

The expression of $K_c$ for above reaction follows:

$K_c=\frac{PCl_5}{PCl_3\times Cl_2}$

We are given:

$[PCl_5]=\frac{5.04\times 10^{-4}mol}{25L}$

$[PCl_3]=\frac{1.60\times 10^{-3}mol}{25L}$

$[Cl_2]=\frac{0.029mol}{25L}$

Putting values in above equation, we get:

$K_c=\frac{(\frac{5.04\times 10^{-4}}{25})}{(\frac{1.60\times 10^{-3}}{25})\times (\frac{0.029}{25})}\\\\K_c=271.6$

Hence, the value of $K_c$ for the given reaction is 271.6

For b:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = 271.6

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $250^oC=250+273=523K$

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=1-2=-1$

Putting values in above equation, we get:

$K_p=271.6\times (0.0821\times 523)^{-1}\\\\K_p=6.32$

Hence, the value of $K_p$ for the reaction is 6.32

7 0

4 years ago

Which ion is this atom most likely to form? Options: a. Na+ b. Na- c. Ne- d. Mg+

SVEN [57.7K]

Answer:

Explanation:

because na + has 11 electrons and has a higher chance to give away one then to gain 7 more. so the answer is a

6 0

4 years ago

Read 2 more answers

Any reaction that absorbs 150 kcal of energy can be classified as

Dennis_Churaev [7]

The answer should be (D) endothermic reaction.

Reason : - Reactions in which energy is absorbed are called endothermic reactions. In your provided question, 150 kcal energy is being absorbed, and thus, we can say that it is an endothermic reaction.

6 0

3 years ago