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2 years ago

Please select the word from the list that best fits the definition The energy of motion

Chemistry

1 answer:

ss7ja [257]2 years ago

3 0

Answer:

Kinetic energy

Explanation:

Objects or particles under motion posses energy in the form of kinetic energy which is directly proportional to the square of the velocity of the given particle.

E= 1/2mv² where m is the mass of the particle and v the velocity of the particle.

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Convert 5.15 meters to inches

sukhopar [10]

Answer:

202.76 inches

Explanation:

1 meter = 39.37 inches

5.15 × 39.37 = 202.7559,when rounded it's 202.76

5 0

2 years ago

Kc = 3.07 x 10-4 at 24°C for 2NOBr(g) ↔ 2NO(g) + Br2(g). If the initial concentration of NOBr = 0.878 M, what is the equilibrium

pav-90 [236]

Answer:

The equilibrium concentration of NO is 0.02124 M.

Explanation:

Given that,

Initial concentration of NOBr = 0.878 M

$k_{c}=3.07\times10^{-4}$

Temperature = 24°C

We know that,

The balance equation is

$2NOBr\Rightarrow 2NO+Br_{2}$

Initial concentration is,

$0.878\Rightarrow 0+0$

Concentration is,

$-2x\Rightarrow 2x+x$

Equilibrium concentration

$0.878-2x\Rightarrow 2x+x$

We need to calculate the value of x

Using formula of concentration

$k_{c}=\dfrac{[NO][Br_{2}]}{[NOBr]^2}$

Put the value into the formula

$3.07\times10^{-4}=\dfrac{[2x][x]}{[0.878-2x]^2}$

$2x^2=3.07\times10^{-4}\times(0.878)^2+3.07\times10^{-4}\times4x^2-2\times2x\times0.878\times3\times10^{-4}$

$2x^2=0.0002367+0.001228x^2-0.0010536x$

$2x^2-0.001228x^2+0.0010536x-0.0002367=0$

$1.998772x^2+0.0010536x-0.0002367=0$

$x=0, 0.01062$

We need to calculate the equilibrium concentration of NO

Using formula of concentration of NO

$concentration\ of\ NO=2x$

Put the value of x

$concentration\ of\ NO=2\times0.01062$

$concentration\ of\ NO=0.02124$

Hence, The equilibrium concentration of NO is 0.02124 M.

7 0

3 years ago

Density measurements were conducted on a 22.5oC sample of water which had a theoretical density of 0.997655 g/ml. A volume of 10

Sophie [7]

Let's divide the three experiments: The experiment with 10.00 mL of water is A), the experiment with 15.00 mL is B), and the experiment with 25.00 mL is C).

(1) Now let's calculate the experimental density of each experiment. Density (ρ) is equal to the mass divided by the volume, thus:

$p_{A} =9.98g/10.00mL=0.998g/mL\\p_{B} =15.61g/15.00mL=1.041g/mL\\p_{C} =25.65g/25.00mL=1.026g/mL$

(2)To calculate the average density, we add each density and divide the result by the number of experiments (in this case 3):

$p_{average}=\frac{p_{1}+p_{2}+p_{3}}{3} \\p_{average}=\frac{(0.998+1.041+1.026)g/mL}{3}\\p_{average}=1.022g/mL$

(3) The percent error is calculated by dividing the absolute value of the substraction of the theorethical and experimental values, by the theoretical value, times 100:

%error= $\frac{|p_{average}-p_{theoretical}|}{p_{theoretical}} *100$

%error= $\frac{|1.022g/mL-0.997655g/mL|}{0.997655g/mL}*100$

%error=2.44 %

7 0

3 years ago

How do I do this<br> I need answers fast

MA_775_DIABLO [31]

1.8 would be the top Sig fig

7 0

2 years ago

2. Which of the following elements does not lose an electron easily? (a) Na (6) F (c) Mg (d) Al

vaieri [72.5K]

Answer:

Answer: (b) F

Explanation:

Sodium has 1, magnesium has 2 and Aluminium has 3 electrons in its outermost shell whereas Fluorine has 7 electrons in its outermost shell hence Fluorine does not lose electrons easily.

The electronic configuration of fluorine is 2,7.

Fluorine is the ninth element with a total of 9 electrons.

The first two electrons will go in the 1s orbital.

The next 2 electrons for F go in the 2s orbital.

The remaining five electrons will go in the 2p orbital. Therefore the F electron configuration will be 1s22s22p5.

3 0

1 year ago