All categories

3 years ago

What is the formula for the aqueous salt produced when hydroiodic acid is neutralized by potassium hydroxide?

Chemistry

1 answer:

coldgirl [10]3 years ago

5 0

Answer:

KOH(aq) + HCI(aq) -----> KCI(aq )+ H2O

base acid salt water

hope this helps :)

Explanation:

You might be interested in

QUESTION 1: Using the experimental data, calculate the standard reduction potential for each metal, and organise them into a sta

suter [353]

Answer:you should search on google

Explanation:

because its useful

6 0

2 years ago

Which gas has approximately the same density as c2h6 at stp no nh3 h20 so2?

zmey [24]

1 mole of any gas occupy 22.4 L at STP (standard temperature and pressure, 0°C and 1 atm).

Let given gases be 1 mole. So their volumes will be the same, 22.4 liters.

Density is the ratio of mass to volume.

By formula; density= mass/volume; d=m/V

To find out masses of gases, do the mole calculation.

By formula; mole= mass/molar mass; n= m/M; m= n*M

Molar masses are calculated as
1. C₂H₆ (ethane) = 2*12 g/mol + 6*1 g/mol= 30 g/mol
2. NO (nitrogen monoxide) = 1*14 g/mol + 1*16 g/mol= 30 g/mol
3. NH₃ (ammonia) = 1*14 g/mol + 3*1 g/mol= 17 g/mol
4. H₂O (water) = 2*1 g/mol + 1*16 g/mol= 18 g/mol
5. SO₂ (sulfur dioxide) = 1*32 g/mol + 2*16 g/mol= 64 g/mol
Use Periodic Table to get atomic mass of elements.

Since their volumes are equal, compounds having the same molar mass will have the same density.
Recall the formula d= m/V.

Ethane and nitrogen monoxide have the same density.

The answer is C₂H₆ and NO.

6 0

2 years ago

In the reaction 2co ( g) + o2( g) → 2co2( g), what is the ratio of moles of oxygen used to moles of co2produced?

KIM [24]

In the reaction 2co ( g) + o2( g) → 2co2( g), the ratio of moles of oxygen used to moles of co2produced is 1:2.

8 0

2 years ago

A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a

deff fn [24]

Answer:

A) $E_{a} = 350KJ/mol$ , $E_{a} = 50KJ/mol$ , $E_{a} = 50KJ/mol$

A = 1.5× $10^{-7}s^{-1}$ , A = 1.9× $10^{-7} s^{-1}$ , A=1.5× $10^{-7} s^{-1}$

B) 4.469

Explanation:

From Arrhenius equation

$K=Ae^{\frac{E_{a} }{RT} }$

where; K = Rate of constant

A = Pre exponetial factor

$E_{a}$ = Activation Energy

R = Universal constant

T = Temperature in Kelvin

Given parameters:

$E_{a} =165KJ/mol$

$T_{1}=505K$

$T_{2}=525K$

$R=8.314JK^{-1}mol^{-1}$

taking logarithm on both sides of the equation we have;

$InK=InA-\frac{E_{a} }{RT}$

since we have the rate of two different temperature the equation can be derived as:

$In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )$

$In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1} }.(\frac{1}{505} -\frac{1}{525} )$

$In(\frac{K_{2} }{K_{1} } )$ = 19846.04×7.544× $10^{-5}$ = 1.497

$\frac{K_{2} }{K_{1} }$ = $e^{1.497}$ = 4.469

6 0

2 years ago

You guys are working hard on this school work... You guys deserve to take a break, come hang out with me.

Wittaler [7]

Answer:

hggggggggggggggggggggggggggggggggggggsssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss

Explanation:

jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj

5 0

2 years ago