Pls help with give brainiest Which statement best explains why the sun's

Some atoms are chemically stable and will not bond with atoms of other

tigry1 [53]

Answer:

The stability of atoms depends on whether or not their outer-most shell is filled with electrons. If the outer shell is filled, the atom is stable. Atoms with unfilled outer shells are unstable, and will usually form chemical bonds with other atoms to achieve stability.

Explanation:

7 0

3 years ago

Read 2 more answers

Need help !!!!! ASAP

Julli [10]

<h2>Hello!</h2>

The answer is:

The new volume will be 1 L.

$V_{2}=1L$

To solve the problem, since we are given the volume and the first and the second pressure, to calculate the new volume, we need to assume that the temperature is constant.

To solve this problem, we need to use Boyle's Law. Boyle's Law establishes when the temperature is kept constant, the pressure and the volume will be proportional.

Boyle's Law equation is:

$P_{1}V_{1}=P_{2}V_{2}$

So, we are given the information:

$V_{1}=2L\\P_{1}=50kPa\\P_{2}=100kPa$

Then, isolating the new volume and substituting into the equation, we have:

$P_{1}V_{1}=P_{2}V_{2}$

$V_{2}=\frac{P_{1}V_{1}}{P_{2}}$

$V_{2}=\frac{50kPa*2L}{100kPa}=1L$

Hence, the new volume will be 1 L.

$V_{2}=1L$

Have a nice day!

4 0

3 years ago

A 50.0 mL sample of an aqueous H2SO4 solution is titrated with a 0.375 M NaOH solution. The equivalence point is reached with 62

vesna_86 [32]

Answer: The concentration of $H_2SO_4$ is 0.234 M

Explanation:

According to the neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity $H_2SO_4$ = 2

$M_1$ = molarity of $H_2SO_4$ solution = ?

$V_1$ = volume of $H_2SO_4$ solution = 50.0 ml

$n_2$ = acidity of $NaOH$ = 1

$M_1$ = molarity of $NaOH$ solution = 0.375 M

$V_1$ = volume of $NaOH$ solution = 62.5 ml

Putting in the values we get:

$2\times M_1\times 50.0=1\times 0.375\times 62.5$

$M_1=0.234M$

Therefore concentration of $H_2SO_4$ is 0.234 M

6 0

3 years ago

the combustion of a sample of butane C4H10 (lighter fluid ),produced 2.16 grams of water. 2C4H10 +13O2 ------->8CO2+10H2O how

ratelena [41]

I think this is learned in chemistry do you have any notes that can help

7 0

3 years ago

The estimated heat of vaporization of diethyl ether using the Chen's rule is A. 29.7 KJ/mol B. 33.5 KJ/mol C. 26.4 KJ/mol D. 36.

Brums [2.3K]

Answer:

C. 26.4 kJ/mol

Explanation:

The Chen's rule for the calculation of heat of vaporization is shown below:

$\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]$

Where,

$\Delta H_v$ is the Heat of vaoprization (J/mol)

$T_b$ is the normal boiling point of the gas (K)

$T_c$ is the Critical temperature of the gas (K)

$P_c$ is the Critical pressure of the gas (bar)

R is the gas constant (8.314 J/Kmol)

For diethyl ether:

$T_b=307.4\ K$

$T_c=466.7\ K$

$P_c=36.4\ bar$

Applying the above equation to find heat of vaporization as:

$\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]$

$\Delta H_v=26400 J/mol$

The conversion of J into kJ is shown below:

1 J = 10⁻³ kJ

Thus,

$\Delta H_v=26.4 kJ/mol$

<u>Option C is correct</u>

6 0

3 years ago