All categories

4 years ago

How did changing from a light drizzle to a downpour affect the river and the sediment un it

Chemistry

1 answer:

soldi70 [24.7K]4 years ago

6 0

Answer:

The velocity of the river increased.

There was more erosion in the stream.

The type of sediment that moved changed.

Explanation:

You might be interested in

How does the acidity of water affect the mass of seashells that have been submerged in it overnight?? I need a balanced chemical

Ulleksa [173]

Answer is: <span>the acidity of water reduce the mass of seashells.
</span>
Balanced chemical reaction:
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l).
Calcium carbonate (CaCO₃) dissolves in hydrochloric acid and gas carbon(IV) oxide is formed.
Calcium carbonate is inorganic salt insoluble in water, but soluble in acidic solutions.

8 0

4 years ago

Match each number to how it would appear in scientific notation.

Verdich [7]

A. 8,330 --> 8.33 × 10^3
B. 83.3 --> 8.33 × 10^1
C. 0.00833 --> 8.33 × 10^-3
D. 0.0833 --> 8.33 × 10^-2
E. 83,330 --> 8.33 × 10^4

Hope this helps!

4 0

3 years ago

Read 2 more answers

A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,

Dafna1 [17]

Answer : The concentration of $A,B\text{ and }C$ at equilibrium are 0.132 M, 0.232 M and 0.168 M respectively.

Explanation :

The given chemical reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

$\Delta G^o=-RT \ln k$

where,

$\Delta G^o$ = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = $25^oC=273+25=298K$

Now put all the given values in the above relation, we get:

$-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k$

$k=5.45$

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

Initially 0.30 0.40 0

At equilibrium (0.30-x) (0.40-x) x

The expression of equilibrium constant will be,

$k=\frac{[C]}{[A][B]}$

$5.45=\frac{x}{(0.30-x)\times (0.40-x)}$

By solving the term x, we get

$x=0.168\text{ and }0.716$

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of $A$ at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M

The concentration of $B$ at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M

The concentration of $C$ at equilibrium = x = 0.168 M

3 0

4 years ago

Balance the equation ___Fe+__O2–>___Fe2O3

Sever21 [200]

Answer:

4 Fe + 3 O2 = 2 Fe2O3

Explanation:

In order to balance the equation, there should be equal number of iron atoms (Fe) and oxygen atoms (O2) in Iron(III) oxide (Fe2O3).

Since there are two atoms of Oxygen in 1 oxygen molecule and 3 atoms of Oxygen in 1 Iron Oxide, the least common multiples of the total oxygen atoms which would equal each other is 6 atoms of Oxygen. This results in 3 molecules of O2 and 2 molecules of Fe2O3.

In order to balance out the amount of iron in iron oxide, you must then calculate the total iron atoms in iron oxide; since there are 2 atoms of Fe in each molecule of Iron Oxide and there is 2 molecules of Fe2O3, the total is 2*2 = 4 atoms of iron, Fe.

In order to get 4 atoms of iron from molecules of Fe, you need 4 molecules of Fe, since each molecule contains 1 Fe.

5 0

3 years ago

Acid rain is closely associated with _______________. select one:

DENIUS [597]

I'm pretty sure its B or C. Sorry, hope this helped some....??

3 0

4 years ago