All categories

3 years ago

URGENT! PLS HELP

Chemistry

1 answer:

dedylja [7]3 years ago

7 0

Given Balanced Reaction :

2N₂ + 3H₂ ====> 2NH₃

The Balanced Reaction tells us that :

When Two Moles of Nitrogen Gas Reacts with 3 Moles of Hydrogen Gas it results in 2 Moles of Ammonia.

Our Question is How many Moles of Ammonia are made when 3 Moles of Hydrogen gas is used.

From the Balanced Reaction it is Obvious that When 3 Moles of Hydrogen gas is used 2 Moles of Ammonia is made.

So, The Answer is 2 Moles of Ammonia is Made.

You might be interested in

Bismuth-210 has a half life of 5.0 days if you had a sample of 60.00g of Bismuth-210 how much would be left after 15 days? (Show

Zielflug [23.3K]

Answer:

7.5 gm left

Explanation:

Bismuth-210 has a half life of 5 days

15 days is 15/5 = 3 half lives

since half the amount is left in 5 days or 1 half life

(1/2) x (1/2) x (1/2) the staring amount would be left in

3 half lives. so 1/8 is left

(1/8) x 60.0 = 7.5 gm left

5 0

3 years ago

BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w

Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

Explanation:

Given: The base dissociation constant: $K_{b}$ = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: $K_{w}$ = 1 × 10⁻¹⁴

The acid dissociation constant ( $K_{a}$ ) for the weak acid (BH⁺) can be calculated by the equation:

$K_{a}. K_{b} = K_{w}$

$\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}$

$\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}$

Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:

Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+

Initial: 0.1 M x x

Change: -x +x +x

Equilibrium: 0.1 - x x x

The acid dissociation constant: $K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}$

$As, x$

$\Rightarrow 0.1 - x = 0.1$

$\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }$

$\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}$

$\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}$

Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

5 0

3 years ago

Mannitol salt agar is defined as both a selective and a differential agar since it selects against non-halophiles and allows the

viktelen [127]

Answer: The above statement is true

Explanation:

Mannitol salt agar is a medium which contain high concentration of salt,about 7.5%-10%. It is commonly used as a selective and differential growth medium.Because of its high concentration of salt,it inhibits growth if non halophiles(as they only grow best in media containing less tha 0.2M salts). If the bacteria can ferment mannitol,an acidic byproduct is formed which turns phenol red agar into yellow.So only those bacteria grow well which can ferment mannitol.

8 0

3 years ago

Which element or elements are unbalanced in this equation?

Stells [14]

Answer:

Only the H is unbalanced.

Explanation:

There are 4 H's and 2 of everything else

7 0

3 years ago

Read 2 more answers

A gas mixture of Ne and Ar has a total pressure of 4.00 atm and contains 16.0 mol of gas. If the partial pressure of Ne is 2.75

Stolb23 [73]

Answer:

5 moles of Argon is present in the mixture.

Explanation:

Total pressure of the gaseous mixture = 4 atm

Total number of moles = 16

Partial pressure of Ne = 2.75 atm

By Dalton's law of partial pressure, the total pressure of gaseous mixture is the sum of partial pressures of individual gases which are non-reactive.

Hence:

$P_{total}=P_{Ar}+P_{Ne}\\4=P_{Ar}+2.75\\P_{Ar}=1.25\ atm$

Also :

Partial pressure = mole fraction*total pressure

$P_{Ar}=X_{Ar}P_{total}$

$X_{Ar}=\frac{1.25}{4}=0.3125$

$\frac{n_{Ar}}{n_{total}}=0.3125\\n_{Ar}=5$

∴Number of moles of Argon = 5

4 0

3 years ago