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2 years ago

PLZZ!!!- I NEED THIS RIGHT NOWA block is pulled 0.90 m to the right in 2.5 s.

Chemistry

1 answer:

UNO [17]2 years ago

4 0

The average speed is 0.28

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Players 1, 2, 3 are playing a tournament. Two of these three players are randomly chosen to play a game in round one, with the w

Anastasy [175]

The answer & explanation for this question is given in the attachment below.

6 0

2 years ago

495 cm3 of oxygen gas and 877 cm3 of nitrogen gas, both at 25.0 C and 114.7 kpa, are injected into an evacuated 536 cm3 flask. F

I am Lyosha [343]

Answer:

Total pressure of the flask is 2.8999 atm.

Explanation:

Given data:

Volume of oxygen (O2) gas= 495 cm3

= 0.495 L (1 cm³ = 1 mL = 0.001 L)

Volume of nitrogen (N2) gas = 877 cm3

= 0.877 L (1 cm³ = 1 mL = 0.001 L)

volume of falsk = 536 cm3

= 0.536 L (1 cm³ = 1 mL = 0.001 L)

Temperature = 25 °C

T = (25°C + 273.15) K

= 298.15 K

Pressure = 114.7 kPa

= 114.700 Pa

Pressure (torr) = 114,700 / 101325

= 1.132 atm

Formula:

PV=nRT (ideal gas equation)

P = pressure

V = volume

R (gas constnt)= 0.0821 L.atm/K.mol

T = temperature

n = number of moles for both gases

Solution:

Firstly we will find the number of moles for oxygen and nitrogen gas.

For Oxygen:

n = PV / RT

n = 1.132 atm × 0.495 L / 0.0821 L.atm/K.mol × 298.15 K

= 0.560 / 24.47

= 0.0229 moles

For Nitrogen:

n = PV / RT

n = 1.132 atm × 0.877 / 0.0821 L.atm/K.mol × 298.15 K

n = 0.992 / 24.47

= 0.0406

Total moles = moles for oxygen gas + moles for nitrogen gas

= 0.0229 moles + 0.0406 moles

n = 0.0635 moles

Now put the values in formula

PV=nRT

P = nRT / V

P = 0.0635 × 0.0821 L.atm/K.mol × 298.15 K / 0.536 L

P = 1.554 / 0.536

P = 2.8999 atm

Total pressure in the flask is 2.8999 atm, while assuming the temperature constant.

7 0

2 years ago

Read 2 more answers

What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)

Delvig [45]

Explanation:

The given data is as follows.

Energy of radiation absorbed by the electron in hydrogen atom = $1.08 \times 10^{-17} J$

As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.

E = $h \nu$

$1.08 \times 10^{-17} J$ = $6.626 \times 10^{-34} Js \times \nu$

$\nu$ = $0.163 \times 10^{17} s^{-1}$

or, $\nu$ = $1.63 \times 10^{16} s^{-1}$

It is known that, $\nu = \frac{c}{\lambda}$

$1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}$

$\lambda$ = $1.84 \times 10^{-8} m$

And, according to De-Broglie equation $\lambda = \frac{h}{p}$

as, p = $m \times \nu$

So, $\lambda = \frac{h}{m \times \nu}$

$m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}$

= $3.6 \times 10^{-26} J/m$

Now, on squaring both the sides we get the following.

$(m \times \nu)^{2}$ = $(3.6 \times 10^{-26} J/m)^{2}$

= $12.96 \times 10^{-52}$

$m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

where, m = mass of electron

So, $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

= $\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}$

= $1.42 \times 10^{-21}$ J

Since, K.E = $\frac{1}{2}m \nu^{2}$

= $\frac{1.42 \times 10^{-21} J}{2}$

= $0.71 \times 10^{-21} J$

Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is $7.1 \times 10^{-22} J$ .

4 0

2 years ago

A 0.15 m solution of chloroacetic acid has a ph of 1.86. What is the value of ka for this acid?

dem82 [27]

Answer: $1.67\times 10^{-3}$

Explanation:

$ClCH_2COOH\rightarrow ClCH_2COO^-+H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Given: c = 0.15 M

pH = 1.86

$K_a$ = ?

Putting in the values we get:

Also $pH=-log[H^+]$

$1.86=-log[H^+]$

$[H^+]=0.01$

$[H^+]=c\times \alpha$

$0.01=0.15\times \alpha$

$\alpha=0.06$

As $[H^+]=[ClCH_2COO^-]=0.01$

$K_a=\frac{(0.01)^2}{(0.15-0.15\times 0.06)}$

$K_a=1.67\times 10^{-3]$

Thus the vale of $K_a$ for the acid is $1.67\times 10^{-3}$

4 0

2 years ago

6.0 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 13

vodomira [7]

Answer:

The molecular formula of X is given as $C_7 H_6 O_3$

Explanation:

$Moles $C O_{2}=\frac{\text { mass }}{\text { molar mass }}=\frac{13.39 \mathrm{g}}{44.01 \mathrm{g} \text { per mole }}=0.304 \mathrm{mol}$\\\\moles $\mathrm{C}=$ moles $\mathrm{CO}_{2}=0.304 \mathrm{mol}$$

$mass $C=$ moles $\times$ molar mass $=0.304 \mathrm{mol} \times 12 \frac{g}{m o l}=3.65g$\\\\moles $\mathrm{H}_{2} \mathrm{O}=\frac{2.35 \mathrm{g}}{18.02 \mathrm{g} \text { permole }}=0.130 \mathrm{mol}$\\\\moles $\mathrm{H}=2 \times$ moles $\mathrm{H}_{2} \mathrm{O}=0.130 \times 2=0.260 \mathrm{mol}$\\\\Mass $\mathrm{H}=0.260 \mathrm{mol} \times 1.008 \frac{g}{\mathrm{mol}}=0.262 \mathrm{g}$$