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2 years ago

What is 680 000 km as an integer?

Chemistry

1 answer:

stellarik [79]2 years ago

7 0

Answer:

INTEGER EXPANDED FORM

❤600000+80000❤

GLAD to HELP

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What is the de Broglie wavelength (in meters) of a 45-g golf ball traveling at 72 m/s?

Cerrena [4.2K]

Answer: The de broglie wavelength is $2.037 \times 10^{-34} m$ .

Explanation:

Calculate $\lambda = \frac{h}{p}$ as follows.

$\lambda = \frac{h}{p}$

where,

h = plank's constant = $6.6 \times 10^{-34} m^{2} kg/s$

p = momentum = $mass \times velocity$

Putting the values in the formula as follows.

$\lambda = \frac{h}{mass \times velocity}$

= $\frac {6.6 \times 10^{-34} m^{2} kg/s}{0.045 kg \times 72 m/s}$

= $2.037 \times 10^{-34} m$

Thus, the de broglie wavelength is $2.037 \times 10^{-34} m$ .

6 0

3 years ago

Read 2 more answers

Well, we might as well try a similar question. Magnesium metal reacts with hydrochloric acid to produce hydrogen gas and an aque

Andrei [34K]

Answer: Magnesium Mg

Explanation:

Oxidization is the process by which a substance either gains oxygen or losses electrons.

The chemical reaction of the above is denoted by,

Mg(s) + 2HCl(aq) -----> MgCl2(aq) + H2(g)

Mg went from a 0 to a +2 state which would mean that it lost electrons.

It was therefore oxidized.

Please do react or comment if you need clarification or if the answer helped you. This can help other users as well. Thank you.

5 0

3 years ago

Provide the structure of the cyanohydrin that results when cyclopentanone reacts with hcn.

dusya [7]

Answer:

All the related answers are attached image formats...Regards

5 0

3 years ago

The law of conservation of matter states that matter cannot be created or destroyed. This law applies when a chemical change occ

lora16 [44]

It says it will never change. So whatever it was before the change, it will be the same after the change. So, 200 should be correct.

4 0

3 years ago

The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure

Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

$P_1=0.69671 atm \approx 0.6970 atm$

The vapor pressure at temperature 363 K is 0.6970 atm

2) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =373 K

$T_2$ = final temperature =383 K

$P_1=1 atm, P_2?$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

The vapor pressure at 383 K is 1.410 atm

8 0

3 years ago