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2 years ago

What is the mass in grams of 6.00 × 10¹⁵ molecules of CS₂?

Chemistry

1 answer:

Elenna [48]2 years ago

3 0

6,160.506

Explanation:

That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 1023 atoms, molecules, or formula units of that substance. In each case, the number of grams in 1 mol is the same as the number of atomic mass units that describe the atomic mass, the molecular mass, or the formula mass, respectively.

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_Na+ Cl2 - -> _NaCl A 2,4 B 1,2 C 3,3 D 2,2

Natali [406]

Answer:

D 2,2

Explanation:

We can see that there are 2 chlorines on the reactant side so there has to be a 2 on the product side

Now we have Na + Cl2 --> 2NaCl

The problem now is that there are 2 sodiums on the product side so add a 2 to the Na on the reactant side

2Na + Cl2 --> 2NaCl

Now it's balanced!

4 0

2 years ago

Write the overall, balanced molecular equation and indicate which element is oxidized and which is reduced for the following rea

galina1969 [7]

Answer:

Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)

Oxidized: Cd

Reduced: Ag

Explanation:

Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)

Cd → Cd²⁺ + 2e⁻ Half reaction oxidation

1e⁻ + Ag⁺ → Ag Half reaction reduction

Ag changed oxidation number from +1 to 0

Cd changed oxidation number from 0 to +2

Let's ballance the electrons

( Cd → Cd²⁺ + 2e⁻ ) .1

( 1e⁻ + Ag⁺ → Ag ) .2

Cd + 2e⁻ + 2Ag⁺ → 2Ag + Cd²⁺ + 2e⁻

Finally the ballance equation is:

Cd(s) + 2AgNO₃(aq) → Cd(NO₃)₂ (aq) + 2Ag(s)

4 0

3 years ago

1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L

anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

Vapor presume of 2‑Propanol $(P^o_2)$ = 45.2 torr

Mole fraction of 1‑Propanol $(x_1)$ = 0.540

Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

where,

$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

$x_1$ = mole fraction of 1‑Propanol

$p_1=(0.540)\times (20.9torr)=11.3torr$

and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

$p_2=(0.46)\times (45.2torr)=20.8torr$

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

$\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352$

and,

$\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648$

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0

3 years ago

Which resource do these turbines gather? A water о в оll C natural gas D. wind

lapo4ka [179]

Answer:

D....

Explanation:

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3 years ago

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NemiM [27]

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2) Is=54· g

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3 years ago