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nexus9112 [7]
3 years ago
15

How does Controlled fires help prevent real forest fires?

Chemistry
1 answer:
Salsk061 [2.6K]3 years ago
6 0

Answer:

it burns things which would be burned easily by lightning, and then people put the fire out immediately.  If the dried plants that they are lighting on fire are hit by lightning, it can lead to a massive fire without anyone realizing.  

You might be interested in
4. The woman's auxiliary at the Fire Company need to make 250 one pound pie crusts for a fund raiser. If
Andre45 [30]

Answer:

i rly did this for the extra points sry btw its d tho

Explanation:

7 0
3 years ago
The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate
Karolina [17]

<u>Answer:</u> The value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

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 = 3.45

K_c = equilibrium constant in terms of concentration = ?

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

T = temperature = 500 K

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

Putting values in above equation, we get:

3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084

The equation used to calculate concentration of a solution is:

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

Initial moles of (CH_3)_3CCl(g) = 1.00 mol

Volume of the flask = 5.00 L

So, \text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M

For the given chemical reaction:

                (CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)

Initial:               0.2                    -                        -

At Eqllm:          0.2 - x               x                       x

The expression of K_c for above reaction follows:

K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}

Putting values in above equation, we get:

0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

[(CH_3)_2C=CH]=x=0.094M

[HCl]=x=0.094M

[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M

Hence, the value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

8 0
3 years ago
A 1.2 kg block of iron at 32 ∘C is rapidly heated by a torch such that 12 kJ is transferred to it. What temperature would the bl
lbvjy [14]

Answer:

For iron

Final temperature = 54,22°C

For copper

Final Temperature = 63.67 °C

Explanation

Hello,

You are using a torch to warm up a block of iron that has an initial temperature of 32°C.

The first you have to know is that the "heat capacity" could simply define as the heat required to go from an initial temperature to a final temperature.

So you need to use the heat capacity equation as follow in the paper.

The equation has to have all terms in the same units, so:

q = 12000 J

s = 0.450 J / g °C

m = 1200 g

Ti = 32 °C

Download odt
3 0
3 years ago
Finding the pH for [H+] = 9.4 * 10-3 M?
saul85 [17]

Answer:

pH = 2.0

Explanation:

To find the pH of a solution, take the -log[H+]. In this case, the -log(9.4 x 10^-3) equals 2.02687 which makes 2.0 when accounting for significant figures.

3 0
3 years ago
For 2,663 kg of a compound with the formula Al(SO), determine the following quantities (4 pts each); a) The number of moles of t
Nastasia [14]

Answer:

a) 35.485 moles of Al(SO)

b) 35.485 moles of S atoms

c) 2.136197(10^{25}) Al atoms

d) 567.723 g of O

Explanation:

Let's define the following terms :

1 mol = 6.02.(10^{23}) elemental units

For example :

1 mol of oxygen atoms = 6.02.(10^{23}) oxygen atoms

Now, our compound has the following formula

Al(SO)

Where Al is aluminium

S is sulfur

And O is oxygen

All the subscripts are 1 so we can say the following :

1 molecule of Al(SO) has 1 atom of Al , 1 atom of S and 1 atom of O

In terms of moles :

1 mol of Al(SO) has 1 mol of Al , 1 mol of S and 1 mol of O

The molar masses of Al, S and O are

molarmass_{(Al)}=26.982\frac{g}{mol}

molarmass_{(S)}=32.065 \frac{g}{mol}

molarmass_{(O)}=15.999\frac{g}{mol}

If we sum all the molar masses =(26.982+32.065+15.999)\frac{g}{mol}=75.046\frac{g}{mol}

Finally, 75.046 g of Al(S0) is 1 mol of Al(SO) which contains 26.982 g of Al, 32.065 g of S and 15.999 g of O.

1 mol of Al(SO) contains 1 mol of Al, 1 mol of S and 1 mol of O.

Now we can calculate a),b),c) and d)

For a)

2.663 kg=2663g

75.046 g of Al(SO) = 1 mol of Al(SO)

2663 g of Al(SO) = x

x=\frac{2663}{75.046}mol=35.485 mol

2.663 kg of Al(SO) contains 35.485 moles of Al(SO)

b) and c) 1 mol of Al(SO) molecules contains 1 mol of S atoms and 1 mol of Al atoms

We have 35.485 moles of Al(SO) molecules so

We have 35.485 moles of S atoms

And 35.485 moles of Al atoms

If 1 mol = 6.02(10^{23})

35.485 moles of Al have (35.485)(6.02)(10^{23})=2.136197(10^{25}) Al atoms

d) 75.046 g of Al(SO) contains 15.999 g of O

2663 g of Al(SO) contains x g of O

x=\frac{(2663).(15.999)}{75.046} g

x = 567.723 g of O

6 0
3 years ago
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