If you were givin 2.50 moles of.sodium bromide, how many grams of salt do you have

Complete the following reaction.<br> 14/7 N + 1/0 n —> <br> blank/blank C+1/1 H

Nataly_w [17]

Answer:

14/6

Explanation:

U 2 can help me by marking as brainliest........

8 0

2 years ago

Read 2 more answers

Use the equation n2 + 3h2=2nh3 if 8.0g N2 react, how many grams of NH3 will be produced

Eduardwww [97]

the balanced equation for the formation of ammonia is

N₂ + 3H₂ ---> 2NH₃

molar ratio of N₂ to NH₃ is 1:2

mass of N₂ reacted is 8.0 g

therefore number of N₂ moles reacted is - 8.0 g / 28 g/mol = 0.286 mol

according to the molar ratio,

1 mol of N₂ will react to give 2 mol of NH₃, assuming nitrogen is the limiting reactant

therefore 0.286 mol of N₂ should give - 2 x 0.286 mol = 0.572 mol of NH₃

therefore mass of NH₃ formed is - 0.572 mol x 17 g/mol = 9.72 g

a mass of 9.72 mol of NH₃ is formed

6 0

3 years ago

3. A 0.025L solution of HCl is neutralized by 0.018L of a 1.0 M NaOH solution. What is the concentration of the HCl solution?

Nookie1986 [14]

The concentration of the HCl solution is 0.72 M.

<h3>How do we calculate the concentration?</h3>

Concentration of the required solution by the use of the known concentration solution will be determine by using the below equation as:

M₁V₁ = M₂V₂, where

M₁ & V₁ are the molarity and volume of the HCl solution.
M₂ & V₂ are the molarity and volume of the NaOH solution.

On putting values in the above equation, we get

M₁ = (1)(0.018) / (0.025) = 0.72 M

Hence required concentration of HCl is 0.72M.

To know kore about molarity, visit the below link:

brainly.com/question/24305514

#SPJ1

4 0

2 years ago

A sample of air is slowly passed through aqueous Sodium hydroxide and then over heater copper. Which gases are removed by this p

bagirrra123 [75]

Carbon dioxide and oxygen are removed from the air.

Explanation:

When air is passed through aqueous sodium hydroxide solution the carbon dioxide is removed from the air.

First the carbon dioxide will dissolve and react with water to form carbonic acid ( H₂CO₃) :

CO₂ + H₂O → H₂CO₃

The the carbonic acid will react with sodium hydroxide to form sodium carbonate (Na₂CO₃):

H₂CO₃ + 2 NaOH → Na₂CO₃ + 2 H₂O

After this by passing the air over heated cooper the oxygen is removed.

2 Cu + O₂ → 2 CuO

Learn more about:

neutralization reaction

brainly.com/question/2632201

#learnwithBrainly

7 0

3 years ago

(6) Compare a CSTR with a PFR below. a. A flow of 0.3 m3/s enters a CSTR (volume of 200 m3) with an initial concentration of spe

Dmitry [639]

Answer:

Explanation:

Given that:

The flow rate Q = 0.3 m³/s

Volume (V) = 200 m³

Initial concentration $C_o$ = 2.00 ms/l

reaction rate K = 5.09 hr⁻¹

Recall that:

$time (t) = \dfrac{V}{Q}$

$time (t) = \dfrac{200}{0.3}$

$time (t) = 666.66 \ sec$

$time (t) = \dfrac{666.66 }{3600} hrs$

$time (t) = 0.185 hrs$

$\text{Using First Order Reaction:}$

$\dfrac{dc}{dt}=kc$

where;

$t = \dfrac{1}{k} \Big( \dfrac{C_o}{C_e}-1 \Big)$

$0.185 = \dfrac{1}{5.09} \Big ( \dfrac{200}{C_e}- 1 \Big)$

$0.942 = \Big ( \dfrac{200}{C_e}- 1 \Big)$

$1+ 0.942 = \Big ( \dfrac{200}{C_e} \Big)$

$\dfrac{200}{C_e} = 1.942$

$C_e = \dfrac{200}{1.942}$

$\mathbf{C_e = 102.98 \ mg/l}$

Thus; the concentration of species in the reactant = 102.98 mg/l

b). If the plug flow reactor has the same efficiency as CSTR, Then:

$t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} =0.196(0.663)$

$\dfrac{V_{PFR}}{0.3 hrs} =0.196(0.663)$

$\dfrac{V_{PFR}}{0.3*3600 sec} =0.196(0.663)$

$V_{PFR} =0.196(0.663)*0.3*3600$

$V_{PFR} = 140.34 \ m^3$

The volume of the PFR is ≅ 140 m³

3 0

3 years ago