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3 years ago

If the atomic number of an element is 9 and the mass number is 19, how many neutrons does the atom have? 10 9 19 28 1 points

1 answer:

6 0

Remember....

mass number= atomic number + number of neutrons

If the mass number is 19 and the atomic number is 9, then the number of neutrons is 19-9 which is 10.

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C. How many moles of gas are in a container with a volume of 9.55 mL at 35 °C and a pressure of 895 mmHg?

olya-2409 [2.1K]

Answer: There are $4.45\times 10^{-4}moles$ of gas are in a container with a volume of 9.55 mL at 35 °C and a pressure of 895 mmHg

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 895 mm Hg= 1.18 atm (760 mm Hg= 1 atm)

V = Volume of gas = 9.55 ml = 0.00955 L (1 L=1000ml)

n = number of moles = ?

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $35^0C=(35+273)K=308K$

$n=\frac{PV}{RT}$

$n=\frac{1.18atm\times 0.00955L}{0.0821L atm/K mol\times 308K}=4.46\times 10^{-4}moles$

Thus there are $4.45\times 10^{-4}moles$ of gas are in a container with a volume of 9.55 mL at 35 °C and a pressure of 895 mmHg

5 0

2 years ago

You made hypochlorous acid (HOCI) by mixing bleach (NaOCI) with:

Annette [7]

<u>Answer:</u> When bleach is mixed with water, it produces hypochlorous acid.

<u>Explanation:</u>

The chemical name for bleach is sodium hypochlorite. When this compound is reacted with water, it produces hypochlorous acid and sodium hydroxide.

The chemical equation for the reaction of sodium hypochlorite and water follows:

$NaOCl+H_2O\rightarrow HOCl+NaOH$

By Stoichiometry of the reaction:

1 mole of sodium hypochlorite reacts with 1 mole of water to produce 1 mole of hypochlorous acid and 1 mole of sodium hydroxide.

Hence, when bleach is mixed with water, it produces hypochlorous acid.

6 0

3 years ago

10 pointsss Where are tsunamis likely to originate?

ycow [4]

Tsunamis are likely to originate in the pacific ocean

6 0

3 years ago

Read 2 more answers

Which type of wave do the particles in a medium move parallel to the direction that the waves move

Talja [164]

Longitudinal waves. In a longitudinal wave the particles in the medium move parallel to the direction waves go. A good example can be the p-waves in an earthquake.

8 0

3 years ago

Use the data given below to construct a Born-Haber cycle to determine the second ionization energy of Ca. Δ H°(kJ) Ca(s)→Ca(g) 1

Drupady [299]

Answer : The value of second ionization energy of Ca is 1010 kJ.

Explanation :

The formation of calcium oxide is,

$Ca(s)+\frac{1}{2}O_2(g)\overset{\Delta H_f}\rightarrow CaO(s)$

$\Delta H_f^o$ = enthalpy of formation of calcium oxide = -635 kJ

The steps involved in the born-Haber cycle for the formation of $CaO$ :

(1) Conversion of solid calcium into gaseous calcium atoms.

$Ca(s)\overset{\Delta H_s}\rightarrow Ca(g)$

$\Delta H_s$ = sublimation energy of calcium = 193 kJ

(2) Conversion of gaseous calcium atoms into gaseous calcium ions.

$Ca(g)\overset{\Delta H_I_1}\rightarrow Ca^{+1}(g)$

$\Delta H_I_1$ = first ionization energy of calcium = 590 kJ

(3) Conversion of gaseous calcium ion into gaseous calcium ions.

$Ca^{+1}(g)\overset{\Delta H_I_2}\rightarrow Ca^{+2}(g)$

$\Delta H_I_2$ = second ionization energy of calcium = ?

(4) Conversion of molecular gaseous oxygen into gaseous oxygen atoms.

$O_2(g)\overset{\Delta H_D}\rightarrow OI(g)$

$\frac{1}{2}O_2(g)\overset{\Delta H_D}\rightarrow O(g)$

$\Delta H_D$ = dissociation energy of oxygen = $\frac{498}{2}=249kJ$

(5) Conversion of gaseous oxygen atoms into gaseous oxygen ions.

$O(g)\overset{\Delta H_E_1}\rightarrow O^-(g)$

$\Delta H_E_1$ = first electron affinity energy of oxygen = -141 kJ

(6) Conversion of gaseous oxygen ion into gaseous oxygen ions.

$O^-(g)\overset{\Delta H_E_2}\rightarrow O^{2-}(g)$

$\Delta H_E_2$ = second electron affinity energy of oxygen = 878 kJ

(7) Conversion of gaseous cations and gaseous anion into solid calcium oxide.

$Ca^{2+}(g)+O^{2-}(g)\overset{\Delta H_L}\rightarrow CaO(s)$

$\Delta H_L$ = lattice energy of calcium oxide = -3414 kJ

To calculate the overall energy from the born-Haber cycle, the equation used will be:

$\Delta H_f^o=\Delta H_s+\Delta H_I_1+\Delta H_I_2+\Delta H_D+\Delta H_E_1+\Delta H_E_2+\Delta H_L$

Now put all the given values in this equation, we get:

$-635kJ=193kJ+590kJ+\Delta H_I_2+249kJ+(-141kJ)+878kJ+(-3414kJ)$

$\Delta H_I_2=1010kJ$

Therefore, the value of second ionization energy of Ca is 1010 kJ.

6 0

2 years ago