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2 years ago

Plz help i’m gonna cry i’m in class rn

Chemistry

1 answer:

iragen [17]2 years ago

3 0

Answer:

if ur gonna cry then just dont cry its simple logic guyss!!!!!!!!!!!!

Explanation:

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Please help make sure its correct thanks

Wewaii [24]

The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)

<h3>Further explanation</h3>

13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :

$\tt mol=\dfrac{mass}{Ar}=\\\\mol=\dfrac{13.02~g}{58.933}\\\\mol=0.221$

Mass of metal iodide formed : 97.12 g, so mass of Iodine :

$\tt =mass~metal~iodide-mass~Cobalt\\\\=97.12-13.02\\\\=84.1~g$

Then mol iodine (MW=126.9045 g/mol) :

$\tt \dfrac{84.1}{126.9045}=0.663$

mol ratio of Cobalt and Iodine in the compound :

$\tt 0.221\div 0.663=1\div 3$

5 0

2 years ago

SOMEONE PLS HELP ME ITS URGENT

Simora [160]

If copper is heated with iron oxide there is no obvious reaction because

copper is less reactive than iron.

On a reactivity chart, copper is far below iron. This makes it impossible for a replacement reaction to occur, so the equation doesn't change.

I hope I helped!

5 0

2 years ago

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Thomas collected the data in Table 1 to answer a statistical question. Which statistical question does the data in Table 1 answe

Nezavi [6.7K]

Answer:

how tall each sibling is

Explanation:

6 0

2 years ago

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When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t

OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

$Mn(s)+2HCl(aq)\rightarrow MnCl_2(aq)+H_2(g)$

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

$m=1.00 g/mL\times 100.0 mL = 100 g$

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

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

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = $4.18 J/^oC$

$T_{final}$ = final temperature = $23.1^oC$

$T_{initial}$ = initial temperature = $28.9^oC$

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

$q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC$

$q=2,242.4 J=2.242 kJ$

Now we have to calculate the enthalpy change during the reaction.

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

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = $\frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol$

$\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol$

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0

2 years ago

When 25.0 mL of a solution of 1.4 x 10-3 M silver nitrate is mixed with 60.0 mL of a solution of 7.5 x 10-4 M sodium chloride a)

LUCKY_DIMON [66]

Step one write the equation for dissociation of AgNO3 and NaCl

that is AgNO3-------> Ag+ + NO3-

NaCl--------> Na+ + Cl-
then find the number of moles of each compound
that is for AgNO3 = ( 1.4 x10^-3 ) x 25/1000= 3.5 x10^-5 moles
Nacl= (7.5 x10^-4)x 60/1000= 4.5 x10^-5 moles

from mole ratio the moles of Ag+= 3.5 x10^-5 moles and that of Cl-= 4.5 x10^-4 moles

then find the total volume of the mixture
that is 25ml + 60 Ml =85ml = 0.085 liters

The Ksp of Agcl = (Ag+) (cl-), let the concentration of Ag+ be represented by x and also the concentration be represented by x

ksp (1.8 x10^-10) is therefore= x^2

find the square root x=1.342 x10^-5

Ag+ in final mixture is = moles of Ag+/total volume - x
that is {(3.5 x10^-5)/0.085} - 1.342 x10^-5=3.98x10^-4

Cl- in the final mixture is =(4.5 x10^-5 /0.085) - 1.342 x10^-5= 5.16 x10^-4

4 0

3 years ago

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