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2 years ago

11

How many grams of carbon dioxide gas are in a 1.0 L balloon at STP?

1 answer:

Bogdan [553]2 years ago

8 0

Answer:

I'm pretty sure its 1.96g

You might be interested in

What is the compound of c4?

klio [65]

Answer:

c4 is an explosive..

contains RDX, DOS, DOA, and PIB.

Explanation:

5 0

2 years ago

How many electrons does (Al) have to give up to achieve a noble gas?

Andre45 [30]

Aluminium has 3 valance electrons in its outer shell. This means that it would need to give up 3 electrons to achieve a full octet.

5 0

3 years ago

Liquid nitrogen has a density of 0.808 g/mL and boils at 77 K. Researchers often purchase liquid nitrogen in insulated 185 L tan

Serga [27]

Answer:

37.14 %

Explanation:

Using the equation, mass, M = D1 * V1

= D2 * V2

Where,

D1 = density of the liquid Nitrogen

D2 = density of gaseous Nitrogen

V1 = volume of the liquid Nitrogen

V2 = volume of the gaseous Nitrogen

Calculating V2,

0.808 * 185 = 1.15 * V2

Volume of Nitrogen after expansion = 129.98 m3.

Volume = L * b * h

= 10 * 10 * 3.5

Volume of the room = 350 m3.

Fraction of air = volume of Nitrogen after expansion/volume of the room * 100

= 129.98/350 *100

= 37.14 %

5 0

3 years ago

If 1.08 g of sodium sulfate reacts with an excess of phosphoric acid, how much sulfuric acid is produced?

vfiekz [6]

Answer: 0.745 g of $H_2SO_4$ will be produced from 1.08 g of sodium sulfate

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Na_2SO_4=\frac{1.08g}{142.04g/mol}=0.0076moles$

$3Na_2SO_4+2H_3PO_4\rightarrow 2Na_3PO_4+3H_2SO_4$

$Na_2SO_4$ is the limiting reagent as it limits the formation of product and $H_3PO_4$ is the excess reagent.

According to stoichiometry :

3 moles of $Na_2SO_4$ produce = 3 moles of $H_2SO_4$

Thus 0.0076 moles of $Na_2SO_4$ will require= $\frac{3}{3}\times 0.0076=0.0076moles$ of $H_2SO_4$

Mass of $H_2SO_4=moles\times {\text {Molar mass}}=0.0076moles\times 98.1g/mol=0.745g$

Thus 0.745 g of $H_2SO_4$ will be produced from 1.08 g of sodium sulfate

3 0

2 years ago

Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0

3 years ago