True or False: A revoked license can be renewed after 10 days.

Why will a toy train remain still if you push on the engine with

netineya [11]

Explanation:

Because when two equal forces are applied from opposite directions, they "eliminate" each other.

The train would go right if a 3N force was only applied in the right direction, and it would go left if the same force was only applied to the left.

If a 3N force was applied to the right and a 2N force to the left, it would equal a 1N force to the right (3-2=1).

But there it's 3-3=0, so in practice the force is 0N. Therefore the train won't move.

6 0

3 years ago

Which of the following is the quantum number set for an electron in the 3rd energy level, dumb-bell shaped orbital, on the z-axi

Juli2301 [7.4K]

Answer:

Explanation:

Principal quantum no "n" = 3

Azimuthal quantum no "l"= 1

Magnetic quantum no "m"= +1/2

Over all is 3pz

5 0

2 years ago

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N

Montano1993 [528]

The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

Answer: The limiting reactant is ammonia $(NH_3)$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For ammonia:

Given mass of ammonia = 135.9 kg = 135900 g (Conversion factor: 1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol$

For carbon dioxide gas:

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol$

The given chemical reaction follows:

$2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)$

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = $\frac{1}{2}\times 7994.12=3997.06mol$ of carbon dioxide

As, given amount of carbon dioxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ammonia is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reactant is ammonia $(NH_3)$

5 0

3 years ago

The equilibrium constant for the gas phase reaction N2 (g) + O2 (g) ⇌ 2NO (g) is Keq = 4.20 ⋅ 10-31 at 30 °C. At equilibrium, __

pychu [463]

Answer:

At equilibrium, reactants predominate.

Explanation:

For every reaction, the equilibrium constant is defined as the ratio between the concentration of products and reactants. Thus, for the reaction N2 (g) + O2 (g) ⇌ 2NO the expression of its equilibrium constant is:

$Keq = \frac{[NO]^{2}}{[O_{2} ][N_{2}]}$

Since the equilibrium constant is Keq = 4.20x10-31 the concentration of reactants O2 and N2 must be much higher than products to obtain such a small number as 4.20x10-31 at the equilibrium. Hence, at equilibrium reactants predominate.

5 0

3 years ago

Read 2 more answers

PLEASE PLEASE HELP!

valentinak56 [21]

Answer: The number of grams of $H_2$ in 1620 mL is 1.44 g