What is the total mass of three diamonds with masses of 14.2 grams, 8.73 grams, and 0.912 grams?

Caluclate the number of grams of oxygen that must react

bonufazy [111]

Answer : The number of grams of oxygen react must be, 170.4 grams.

Solution : Given,

Mass of $C_3H_8$ = 46.85 g

Molar mass of $C_3H_8$ = 44 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_3H_8$ .

$\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{46.85g}{44g/mole}=1.065moles$

Now we have to calculate the moles of $O_2$

The balanced chemical reaction is,

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From the balanced reaction we conclude that

As, 1 mole of $C_3H_8$ react with 5 mole of $O_2$

So, 1.065 moles of $C_3H_8$ react with $1.065\times 5=5.325$ moles of $O_2$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(5.325moles)\times (32g/mole)=170.4g$

Therefore, the number of grams of oxygen react must be, 170.4 grams.

8 0

3 years ago

On another planet, the isotopes of titanium have the given natural abundances. Isotope Abundance Mass (amu) 46Ti 72.000% 45.9526

tiny-mole [99]

Answer:

46.784886 amu is the average atomic mass of titanium on that planet.

Explanation:

Fractional abundance of Ti-46 =72.000% = 0.72000

Atomic mass of Ti-46 = 45.95263 amu

Fractional abundance of Ti-48 =14.300% = 0.14300

Atomic mass of Ti-48 = 47.94795 amu

Fractional abundance of Ti-50 =13.700% = 0.13700

Atomic mass of Ti-50 = 49.94479 amu

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Fractional abundance)}_i\times \text{(Atomic mass of an isotopes})_i$

Average atomic mass of titanium atom:

$0.72000\times 45.95263 amu+0.14300\times 47.94795 amu+0.13700\times 49.94479 amu=46.784886 amu$

46.784886 amu is the average atomic mass of titanium on that planet.

3 0

2 years ago

As a result of the particles in a gas being in constant motion, gas has a _______.

sergejj [24]

Answer:

i think it's variable pressure

if not soo advance sorry :)

7 0

2 years ago

Se realiza una mezcla de minerales de Cu y Fe: 20 kg FeS2 (pirita), 70 kg de Fe2O3 (hemetita) 15 kg de CuFe2 (calcopirita) y 90

Artemon [7]

Answer:

34.78% Fe

39.66% Cu

5.48% S

20.07% O

Explanation:

Para resolver esta pregunta debemos hallar la masa de cada átomo en cada mineral. Así, podremos hallar el porcentaje de cada átomo:

Pirita (Fe: 55.845g/mol; S: 32.065g/mol; FeS2: 119.975g/mol)

Masa Fe:

20kg FeS2 * (1*55.845g/mol / 119.975g/mol) = 9.31kg Fe

Masa S:

20kg FeS2 * (2*32.065g/mol / 119.975g/mol) = 10.69kg S

Hemetita (Fe: 55.845g/mol; O: 16g/mol; Fe2O3: 159.688g/mol)

Masa Fe:

70kg Fe2O3 * (2*55.845g/mol / 159.688g/mol) = 48.96kg Fe

Masa O:

70kg Fe2O3 * (3*16g/mol / 159.688g/mol) = 21.04kg O

Calcopirita (Fe: 55.845g/mol; Cu: 63.546g/mol; CuFe2: 175.236 g/mol)

Masa Fe:

15kg CuFe2 * (2*55.845g/mol / 175.236 g/mol) = 9.56kg Fe

Masa Cu:

15kg CuFe2 * (1*63.546g/mol / 175.236 g/mol) = 5.44kg Cu

Tenorita (O: 16g/mol; Cu: 63.546g/mol; CuO: 79.545 g/mol)

Masa O:

90kg CuO * (1*16g/mol / 79.545 g/mol) = 18.10kg O

Masa Cu:

90kg CuO * (1*63.546g/mol / 79.545 g/mol) = 71.90kg Cu

Masa Total: 20kg + 70kg + 15kg + 90kg = 195kg

Porcentaje Hierro:

9.31kg Fe + 48.96kg Fe + 9.56kg Fe / 195kg * 100 =

34.78% Fe

Porcentaje Cobre:

5.44kg Cu + 71.90kg Cu / 195kg * 100 =

39.66% Cu

Porcentaje Azufre:

10.69kg S / 195kg * 100 =

5.48% S

Porcentaje Oxígeno:

21.04kg O + 18.10kg O/ 195kg * 100 =

20.07% O

8 0

2 years ago

Calculate the number of molecules in 1500 mL of a gas measured under a pressure of 800.0

k0ka [10]

Answer:

3.97 x 10²² molecules

Explanation:

To find the moles, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (0.08206 atm*L/mol*K)

-----> T = temperature (K)

Before you can plug the values into the equation, you need to (1) convert the pressure from torr to atm (by dividing by 760), then (2) convert the volume from mL to L (by dividing by 1000), and then (3) convert the temperature from Celsius to Kelvin (by adding 273).

P = 800.0 torr / 760 = 1.05 atm R = 0.08206 atm*L/mol*K

V = 1500 mL / 1000 = 1.5 L T = 19.0 °C + 273 = 292 K

n = ? moles

PV = nRT

(1.05 atm)(1.5 L) = n(0.08206 atm*L/mol*K)(292 K)

1.579 = n(23.96)

0.0659 = moles

Now, you need to convert moles to molecules using Avogadro's Number.

Avogadro's Number:

6.022 x 10²³ molecules = 1 mole

0.0659 moles 6.022 x 10²³ molecules
------------------------ x ------------------------------------- = 3.97 x 10²² molecules
1 mole

5 0

1 year ago