How many grams of chlorine gas are needed to

The atomic number of an element is the total number of which particles in the nucleus?

VARVARA [1.3K]

An atom of element has there subatomic particles namely, proton, electron and neutron. Here, for a neutral atom, number of proton is equal to number of electron (this is not in the case of ions), this is equal to atomic number of an atom. In an atom, nucleus contains protons and neutrons which is responsible for mass of the atom and electrons move around nucleus in fixed orbits. Thus, atomic mass of an atom is equal to sum of number of protons and neutrons.

Option (b): Proton is the particle in nucleus of an atom, whose total number is equal to atomic number of that atom.

(4) Option (b): Atoms of same element have same atomic number because mass number can be different for different isotopes of atom. Since, atomic number is equal to number of protons, thus, number of protons are same for all atoms of the same element.

(5) Option (d): Isotopes are defined as atoms of same element with same atomic number but different mass number. Thus, correct option is (d) mass numbers.

4 0

3 years ago

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A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

3 years ago

An atom has a mass number of 43 and an<br> atomic number of 23. How many protons<br> does it have?

Airida [17]

Answer:

23

Explanation:

8 0

3 years ago

Choose the best answer please .

STALIN [3.7K]

Answer:

B

Explanation:

3 0

3 years ago

How would I do this? A beaker contains 50.0 mL of 0.25 M aluminum nitrate solution. What is the minimum volume of 0.2 M sodium s

Serhud [2]

The minimum volume of 0.2M sodium sulfide that will precipitate out aluminum from 50.0 mL, 0.25 M aluminum nitrate would be 0.094 L or 94 mL

<h3>Stoichiometric calculation</h3>

From the equation of the reaction:

$2Al(NO_3)_3 + 3Na_2S ---> Al_2S_3 + 6NaNO_3$

Mole ratio of Na2S and Al(NO3)3 = 3:2

Mole of 50.0 mL, 0.25 M Al(NO3)3 = 50/1000 x 0.25

= 0.0125 mole

Equivalent mole of Na2S = 3/2 x 0.0125

= 0.0188 mole

Volume of 0.2M, 0.0188 mole Na2S = 0.0188/0.2

= 0.094 L or 94 mL

More on stoichiometric calculations can be found here: brainly.com/question/8062886

6 0

2 years ago