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expeople1 [14]
3 years ago
6

Ri

Chemistry
1 answer:
trasher [3.6K]3 years ago
6 0

Answer:

Ammonia is a compound.

Explanation:

A compound is composed of two or more separate elements. Ammonia is NH3, which happens to be two different elements.

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Give three examples,from the lab where potential energy was converted to kinetic energy?​
creativ13 [48]

Answer:

*a car geared neutral that is at a stop. two men start pushing it

*tire physical therapy. trie is at rest until you come up to the tire and flip it

*pushing a wheel barrel  

Explanation:

everthing has to be at a complete stop then through energy be moved

7 0
3 years ago
Enyzmes change the _______ of a reaction by making the reaction go ______
11111nata11111 [884]
Rate, fast :Maybe these are the answers.
6 0
3 years ago
An herbicide contains only C, H, Cl, and N. The complete combustion of a 200.0 mg sample of the herbicide in excess oxygen produ
MAVERICK [17]

Answer:

%C = 56,1%

%H = 5,5%

%Cl = 27,6%

%N = 10,8%

Explanation:

The moles of CO₂ are the same than moles of C in the herbicide.

Moles of H₂O are ¹/₂ of moles of H in the herbicide.

Moles of CO₂ are obtained using:

n = PV/RT

Where, in STP: P is 1 atm; V is 0,2092L; R is 0,082atmL/molK; T is 273 K

moles of CO₂ are: 9,345x10⁻³ mol≡ mol of C×12,01g/mol = <em>0,1122 g of C ≡ 112,2mg of C</em>

In the same way, moles of H₂O are 5,450x10⁻³mol×2 =0,1090 mol of H×1,01g/mol = <em>0,0110 g of H ≡ 11,0mg of H</em>

As you have 55,14 mg of Cl, the mg of N are:

200,0mg - 112,2 mg of C - 11,0 mg of H - 55,14 mg of Cl = 21,66 mg of N

Thus, precent composition of the herbicide is:

%C = \frac{112,2 mgC}{200,0mg}×100 = 56,1%C

%H = \frac{11,0 mgH}{200,0mg}×100 = 5,5%H

%Cl = \frac{55,14 mgCl}{200,0mg}×100 = 27,6%Cl

%N = \frac{21,66 mgN}{200,0mg}×100 = 10,8%N

I hope it helps!

3 0
3 years ago
At what temperature would 2.10moles of N2 gas have a pressure of 1.25atm and fill a 25.0 L tank
hodyreva [135]

Answer:

\large \boxed{\text{-92 $^{\circ}$C}}

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data  

p = 1.25 atm

V = 25.0 L

n = 2.10 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

1. Temperature in kelvins

\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}

2. Temperature in degrees Celsius

\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}

8 0
3 years ago
The pressure of a sample of argon gas was increased from 3.14 atm to 7.98 at a constant temperature. If the final volume of argo
noname [10]

Answer:

<h2>36.09 L</h2>

Explanation:

The initial volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume.

Since we're finding the initial volume

V_1 =  \frac{P_2V_2}{P_1}  \\

We have

V_1 =  \frac{7.98 \times 14.2}{3.14} =   \frac{113.316}{3.14}  \\  = 36.0878...

We have the final answer as

<h3>36.09 L</h3>

Hope this helps you

8 0
2 years ago
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