All categories

3 years ago

Graph B

Chemistry

1 answer:

Likurg_2 [28]3 years ago

8 0

Answer:

graph B

As it turns out, FINA used to. In 1972, Sweden's Gunnar Larsson beat American Tim McKee in the 400m

individual medley by 0.002 seconds. That finish led the governing body to eliminate timing by a

significant digit. But why?

You might be interested in

In the electron cloud model, the electron cloud is denser in some locations than in others.

Sonbull [250]

<span>In the electron cloud model, the denser areas represent that there is a great probability that a good number of electrons are ganged up or crowded in that area. The electrons affect the density of some parts of the electron cloud when they condense in those locations.</span>

8 0

3 years ago

I need help with B and C

lakkis [162]

B) Mg is the alkaline earth metal w/12 protons so following the periodic table to the halogen in the same period is

Cl: Chlorine

C) The Neutral noble has w/ 18 electrons is argon so the metal in the same row is

Na: Sodium

8 0

3 years ago

Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $137ml+137ml=274ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 274 g

$T_{final}$ = final temperature of water = 325.8 K

$T_{initial}$ = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

$q=274g\times 4.18J/g^oC\times (325.8-298)K$

$q=31839.896J=31.84KJ$

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

$\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0

4 years ago

Can Someone help me pls?

Ilia_Sergeevich [38]

Answer:

For H3O concentration you do 10^-pH so if pH is 5 then H3O+ is 10^-5= 1*10^-5 H3O+ ions

For OH is one extra step. First find H3o+ ions using equation above then you have to use that to divide 1*10^-14

So if pH is 5....the H3O+ is 1*10^-5 then OH- = (1*10^-14)/(1*10^-5) = 1*10^-9 OH ions

as far as acid/base pH 0-6 is Acid 8-14 is Base. pH of 7 is neutral. Recheck your work *hint* *hint* water is neutral. Spit is above 7 so is base.

8 0

3 years ago

One of the reactions in a blast furnace used to reduce iron is shown above. How many grams of Fe2O3 are required to produce 15.5

Salsk061 [2.6K]

Answer : The correct option is, (b) 22.1 g

Solution : Given,

Mass of iron = 15.5 g

Molar mass of iron = 56 g/mole

Molar mass of $Fe_2O_3$ = 160 g/mole