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3 years ago

In location B, what can you conclude about the sediments carried in the river?

Chemistry

2 answers:

Rama09 [41]3 years ago

5 0

They can form a delta, because the river location B leads into the Gulf of Mexico.

valentina_108 [34]3 years ago

3 0

They can cause it to flood...

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Select the correct answer. Which statement is true for a heating curve?

Anon25 [30]

Answer:

Explanation:

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2 years ago

Which of the following is an example of physical change

BartSMP [9]

A physical change is like freezing water and it turns into Ice.

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2 years ago

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A machine has a mechanical advantage of 0.6. What force should be applied to the machine to make it apply 600 N to an object?

Vinil7 [7]

Answer:

1000N is needed to be applied.

Explanation:

Machines make doing work easier. They allow us use small effort to carry out work on huge amount of load.

The mechanical advantage of a machine;

(M.A) =load/effort

M.A = 0.6

Load =600N

effort =?

0.6 = 600/effort

effort = 600/0.6

effort = 1000N

8 0

3 years ago

calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units

cricket20 [7]

Answer:

21.2 gm

Explanation:

calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units

butane is the hydrocarbon C4H10

in combustion, we react hydrocarbons with O2 to form CO2 and H2O

C4H10 + O2----------------> CO2 + H2O

BALANCE

2C4H10 + 1302--------> 8CO2 + 10 H2O

the molar mass of CO2 is 12 + 16X2 = 44

64.1 gm of CO2 is

64.1/44 = 1.46 MOLES OF CO2,

FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE IT IS A

8:2 OR 4:1 RATIO. THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2

THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES

THE MOLAR MASS OF BUTANE IS 58.12

0.365 MOLES OF C4H10 HAS A MASS OF 0.365 X 58.12 = 21.2 gm

6 0

2 years ago

A reactant decomposes with a half-life of 29.5 s when its initial concentration is 0.229 M. When the initial concentration is 0.

Dmitrij [34]

Answer :

The order of reaction is, 0 (zero order reaction).

The value of rate constant is, $0.00388Ms^{-1}$

Explanation :

Half life : It is defined as the time in which the concentration of a reactant is reduced to half of its original value.

The general expression of half-life for nth order is:

$t_{1/2}\propto \frac{1}{[A_o]^{n-1}}$

or,

$\frac{t_{1/2}_1}{t_{1/2}_2}=\frac{[A_2]^{n-1}}{[A_1]^{n-1}}$

or,

$n=\left(\frac{\log\frac{(t_{1/2})_1}{(t_{1/2})_2}}{\log\frac{(A)_2}{(A)_1}}\right )+1$ .............(1)

where,

$t_{1/2}$ = half-life of the reaction

n = order of reaction

[A] = concentration

As we are given:

Initial concentration of A = 0.229 M

Final concentration of A = 0.639 M

Initial half-life of the reaction = 29.5 s

Final half-life of the reaction = 82.3 s

Now put all the given values in the above formula 1, we get:

$n=\left (\frac{\log \frac{29.5}{82.3}}{\log\frac{0.639}{0.229}}\right )+1$

$n=0.000196\approx 0$

Thus, the order of reaction is, 0 (zero order reaction).

Now we have to determine the rate constant.

To calculate the rate constant for zero order the expression will be:

$t_{1/2}=\frac{[A_o]}{2k}$

When,

$t_{1/2}$ = 29.5 s

$[A_o]$ = 0.229 M

$29.5s=\frac{0.229M}{2k}$

$k=0.00388Ms^{-1}$

Thus, the value of rate constant is, $0.00388Ms^{-1}$

4 0

3 years ago