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Sonbull [250]
3 years ago
14

Assessment timer and count

Chemistry
2 answers:
ollegr [7]3 years ago
8 0

Answer:

The answer to your question would be "the insect touching the trigger hairs".

Explanation:

I'm not much of an ex-plainer but I know this is the right answer because I took the test and got a 100%. Please trust me on this. If wrong, please tell me. This is what I was taught at school. Thank you and good day.

love history [14]3 years ago
4 0

Answer:

The answer is: "the insect touching the trigger hairs"

Explanation:

I took the k-12 test and got a 100%!

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State which separation method you would use to separate potassium iodine solution​
navik [9.2K]
Could be separated by distillation.
3 0
2 years ago
Suppose that coal of density 1.5 g/cm^3 is pure carbon. (It is, in fact, much more complicated, but this is a reasonable first a
NISA [10]

Answer:

q = -6464.9 kJ

Explanation:

We are given that the heat of combustion is  ∆H° = −394 kJ per mol of carbon.Therefore what we need to do is calculate how many moles of C are in the lump of coal by finding its mass since the density is given.

vol = 5.6 cm x 5.1 cm x 4.6 cm = 131.38 cm³

m = d x v = 1.5 g/cm³ x 131.38 cm³ = 197.06 g

mol C = m/MW = 197.06 g/ 12.01g/mol = 16.41 mol

q =  −394 kJ /mol C x 16.41 mol C = -6464.9 kJ

7 0
3 years ago
During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce
swat32

Answer: 16.32 g of O_2 as excess reagent are left.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol

\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)  

According to stoichiometry :

2 moles of SO_2 require = 1 mole of O_2

Thus 0.34 moles of SO_2 will require=\frac{1}{2}\times 0.34=0.17moles  of O_2

Thus SO_2 is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

Moles of O_2 left = (0.68-0.17) mol = 0.51 mol

Mass of O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g

Thus 16.32 g of O_2 as excess reagent are left.

3 0
3 years ago
How many grams of chlorine gas are present in a 150. liter cylinder of chlorine held at a pressure of 1.00 atm and 0. °C? Group
OlgaM077 [116]

Answer:

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

  • P= 1.00 atm
  • V= 150 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 0 C= 273 K

Replacing:

1.00 atm* 150 L= n*0.08206 \frac{atm*L}{mol*K} *273 K

Solving:

n=\frac{1.00 atm* 150 L}{0.08206 \frac{atm*L}{mol*K}*273 K}

n= 6.69 moles

Being Cl= 35.45 g/mole, the molar mass of chlorine gas is:

Cl₂=2*35.45 g/mole= 70.9 g/mole

So if 1 mole has 70.9 grams, 6.69 moles of the gas, how much mass does it have?

mass=\frac{6.69 moles*70.9 grams}{1 mole}

mass= 474.321 grams ≅ 474 grams

<u><em>474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C</em></u>

4 0
3 years ago
Why California and Nevada have so many faults?
gtnhenbr [62]

Answer:

They are due to past earthquakes

6 0
3 years ago
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