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Julli [10]
1 year ago
10

What is the number of moles of 0.0960g of H2SO4

Chemistry
1 answer:
Vilka [71]1 year ago
5 0
<span>Answer</span>=.000978802802moles H2SO4

How I Got My Answer

<span>Molar mass 
</span>H2SO4= 98.079g/mol

What I have
.0960g H2SO4

Equation
.0960g*1mol/98.079g= .000978802802mol H2SO4


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An equimolar mixture of N2(g) and Ar(g) is kept inside a rigid container at a constant temperature of 300 K. The initial partial
andrey2020 [161]

Answer:

The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm

Explanation:

Using the ideal gas law

PV=nRT

if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n . Therefore

Inicial state ) P₁V=n₁RT

Final state )  P₂V=n₂RT

dividing both equations

P₂/P₁ = n₂/n₁ → P₂= P₁ * n₂/n₁

now we have to determine P₁ and n₂ /n₁.

For P₁ , we use the Dalton`s law , where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)

p ar₁ = P₁ * x ar₁  →  P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm

n₁ = n ar₁ + n N₁ =  n ar₁ + n ar₁ = 2 n ar₁

n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁

n₂ /n₁ = 3/2

therefore

P₂= P₁ * n₂/n₁ = 1.5 atm * 3/2  = 2.25 atm

P₂= 2.25 atm

8 0
1 year ago
Write a net ionic equation to show how piperidine, c5h11n, behaves as a base in water.
lidiya [134]
C_5H_{11}N + H2O ---\ \textgreater \  C_5H_{11}NH^{+} + OH^{-}

Nitrogen lone pair will act as a base,removing H+ from water leaving behind OH- ion. 
Why ?
Because N is a better donor than O.
7 0
2 years ago
Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:(a) |(b) Sr(c) K(d) N(e) S(f)
Pavel [41]

Answer:

(a) I⁻ (charge 1-)

(b) Sr²⁺ (charge 2+)

(c) K⁺ (charge 1+)

(d) N³⁻ (charge 3-)

(e) S²⁻ (charge 2-)

(f) In³⁺ (charge 3+)

Explanation:

To predict the charge on a monoatomic ion we need to consider the octet rule: atoms will gain, lose or share electrons to complete their valence shell with 8 electrons.

(a) |

I has 7 valence electrons so it gains 1 electron to form I⁻ (charge 1-).

(b) Sr

Sr has 2 valence electrons so it loses 2 electrons to form Sr²⁺ (charge 2+).

(c) K

K has 1 valence electron so it loses 1 electron to form K⁺ (charge 1+).

(d) N

N has 5 valence electrons so it gains 3 electrons to form N³⁻ (charge 3-).

(e) S

S has 6 valence electrons so it gains 2 electrons to form S²⁻ (charge 2-).

(f) In

In has 3 valence electrons so it loses 3 electrons to form In³⁺ (charge 3+).

3 0
2 years ago
A scientist is wondering why a certain region in the ocean doesn't have maximum phytoplankton growth, despite having plenty of n
Advocard [28]

Answer:

C The water had adequate nitrogen and phosphorus, so it is likely iron limited.

Explanation:

Phytoplankton are single- cell organisms that live in oceans.

They require nitrogen, phosphorus and trace amount of iron to survive.

From the scientist's results after testing the water for nitrogen and phosphorus,there are reasonable amount of these elements.

Therefore insufficient iron in the water is the reason why he could find plenty phytoplankton in the ocean.

8 0
2 years ago
A magnesium ion, Mg2+, with a charge of 3.2×10−19C and an oxide ion, O2−, with a charge of −3.2×10−19C, are separated by a dista
baherus [9]

Explanation:

Formula for work done is as follows.

           W = -k \frac{q_{1}q_{2}}{d}    

where,  k = proportionality constant = 8.99 \times 10^{9} Jm/C^{2}

            q_{1} = charge of Mg^{2+} = 3.2 \times 10^{-19} C

            q_{2} = charge of O_{2-} = -3.2 \times 10^{-19} C

            d = separation distance = 0.45 nm = 0.45 \times 10^{-9} m

Now, we will put the given values into the above formula and calculate work done as follows.

         W = -k \frac{q_{1}q_{2}}{d}    

           = \frac{-[8.99 \times 10^{9} Jm/C^{2} \times 3.2 \times 10^{-19} C \times -3.2 \times 10^{-19} C]}{0.25 \times 10^{-9} m}  

           = 3.68 \times 10^{-18} J

Thus, we can conclude that work required to increase the separation of the two ions to an infinite distance is 3.68 \times 10^{-18} J.

7 0
2 years ago
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