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3 years ago

Which method used to attack ground cover fires is generally used against fires that are very hot, very big, and very fast?

Chemistry

1 answer:

AURORKA [14]3 years ago

6 0

Answer:

Indirect attack (ground cover)

Explanation:

Indirect attack (ground cover) is a method of controlling a groundcover fire whereby the control line is erected or situated at a remote from the main fire edge, and the fuel between the two points was burned.

The indirect attack is set up at some distance

from the raging fire. The indirect attack method is normally used to put out

fires that are too fast, too big or too hot for a direct attack method to work.

adjustments can be made when required

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siniylev [52]

Answer:

Explanation:

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3 years ago

. Explaining How is the legislative branch organized, and what are its functions?

Fed [463]

Answer:

The most important function of the legislative branch is its lawmaking authority. In order for a law to be created, a bill must be introduced by either a member of the House or Senate. Once introduced, the bill is brought to a committee for review. ... Each committee is organized around a specific policy function.

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3 years ago

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

Answer: The freezing point of solution is 2.6°C

Explanation:

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$

Hence, the freezing point of solution is 2.6°C

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3 years ago

6. Cuando se oxidan en el aire 12,120 g de vapor de Zinc se obtienen 15,084 g del óxido. ¿Cuál es la fórmula empírica del óxido?

Eddi Din [679]

Answer:

12120 g + O2 = 15084 g

m Zn = 12.120 Kg

m óxido = 15.084 Kg

1. calcular la masa de cinc en gramos

g = 12,120 Kg x 1000 = 12120 g de cinc

g = 15.084 Kg x 100 = 15084 g de oxígeno

2. calcular gramos de Oxigeno

g O = 15084 g - 12120 g = 2964 g O2

3. calcular % de Zn y O

%m/m ( m soluto / m solc.) x 100

%m/m (Zn) = ( 1210 g / 15084 g ) x 100

% m/m (Zn) = 80.35 % = 80.35 g

%m/m (O) = ( 2964 g / 15084 g ) x 100

% m/m (Zn) = 19.65 % = 19.65 g

4. Calcular moles de cada elemento

Zn: 80.35 g / 65.38 g/mol = 1.228 mol

O: 19.65 g / 16 g/mol = 1.228 mol

5. dividir entre el menor de los elementos

Zn: 1.228 mol / 1.228 mol = 1

O: 1.228 mol / 1.228 mol = 1

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3 years ago

Do you think it matters if you

vovikov84 [41]

Answer:

yes I do think they mean the same thing

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150 grams and 150 grams is the same thing and adding 0 to the end of a decimal does not change its value, you could even put 150.0000000 grams and it would still be equivalent to the other numbers

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2 years ago