What is the atomic mass of an atom that has 6 protons, 6 neutrons, and 6 electrons?

Which statements describing chemical and nuclear reactions are true?

Arte-miy333 [17]

Answer:

What statements?

Explanation:

they both release harmful chemicals and can pollute the earth and destroy our ecosystems.

8 0

3 years ago

(Please help will mark brainliest!)

arsen [322]

We know that Weighted atomic mass of Element is Calculated based upon their existence of isotopes and the Relative abundance of these isotopes.

Given that the Element X is Composed of :

Isotope Relative abundance Atomic Mass

⁵⁵X 70% 55

⁵⁶X 20% 56

⁵⁷X 10% 57

Weighted Atomic Mass of Element X :

= (0.70)(55) + (0.20)(56) + (0.10)(57)

= 38.50 + 11.2 + 5.70

= 55.4

So, the Weighted Atomic Mass of Element X is 55.4

6 0

3 years ago

In a redox reaction, how does the total number of electrons lost by the oxidized substance compare to the total number of electr

snow_lady [41]

The answer is (2) equal to. In redox reactions, you can't just lose electrons somewhere. If an electrons is lost by one, it must be gained by another. Hence, the importance of balancing redox reactions.

3 0

3 years ago

Can someone help balance this?

FromTheMoon [43]

Explanation:

2Fe+3Br---> 2FeBr3

it is balanced this way

8 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