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3 years ago

Based on what you’ve determined regarding one mole of sand and one mole of water, which statements must be correct?

Chemistry

1 answer:

Art [367]3 years ago

8 0

Answer:

I dont know the anwer but i have the same problem

Explanation:

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Competition occurs when

Katen [24]

Answer:

Competition occurs when <em>(A) two or more organisms need the same resource.</em>

Explanation:

I dont think it could be (B), (C) or (D) because those don't really make sense.

3 0

2 years ago

In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced b

kirill115 [55]

Answer:

A) 0.065 M is its molarity after a reaction time of 19.0 hour.

B) In 52 hours $[Co(NH_3)5Br]^{2+}$ will react 69% of its initial concentration.

Explanation:

$Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)$

The reaction is first order in $[Co(NH_3)5Br]^{2+}$ :

Initial concentration of $[Co(NH_3)5Br]^{2+}$ = $[A_o]=0.100 M$

a) Final concentration of $[Co(NH_3)5Br]^{2+}$ after 19.0 hours= $[A]$

t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)

Rate constant of the reaction = k = $6.3\times 10^{-6} s^{-1}$

The integrated law of first order kinetic is given as:

$[A]=[A_o]\times e^{-kt}$

$[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}$

$[A]=0.065 M$

0.065 M is its molarity after a reaction time of 19.0 h.

Initial concentration of $[Co(NH_3)5Br]^{2+}$ = $[A_o]=x$

Final concentration of $[Co(NH_3)5Br]^{2+}$ after t = $[A]=(100\%-69\%) x=31\%x=0.31x$

Rate constant of the reaction = k = $6.3\times 10^{-6} s^{-1}$

The integrated law of first order kinetic is given as:

$[A]=[A_o]\times e^{-kt}$

$0.31x=x\times e^{-6.3\times 10^{-6} s^{-1}\t}$

t = 185,902.06 s = $\frac{185,902.06 }{3600} hour = 51.64 hours$ ≈ 52 hours

In 52 hours $[Co(NH_3)5Br]^{2+}$ will react 69% of its initial concentration.

6 0

3 years ago

Question 4

vazorg [7]

Answer:

H(aq) + NO3 (aq) + HF(aq)

Explanation:

In the given mixture of HNO3 (Nitric Acid) and HF (hydrofluoric acid) in water the major species present are H(aq) + NO3 (aq) + HF(aq).

On the reaction of HNO3 (Nitric Acid) and HF (hydrofluoric acid) in water , it will give a polar solution and will form a homogenous mixture.

Hence, the correct answer is "H(aq) + NO3 (aq) + HF(aq)".

4 0

3 years ago

Round each number to four significant figures

ira [324]

Answer:

84,791 » 8479

256.75 » 256.8

431,801 » 4318

0.00078100 » 0.0007810

Explanation:

$.$

5 0

3 years ago

Calculate the mole fraction of cai2 in an aqueous solution prepared by dissolving 0.400 moles of cai2 in 850.0 g of water.

alukav5142 [94]

1) Formulas:

a) mole fraction of component 1, X1

X1 = number of moles of compoent 1 / total number of moles

b) Molar mass = number grams / number of moles => number of moles = number of grams / molar mass

2) Application

Number of moles of CaI2 = 0.400

Molar mass of water = 18.0 g/mol

Number of moles of water: 850.0 g / 18.0 g/mol = 47.22 mol

Total number of moles = 0.400 + 47.22 =47.62

Molar fraction of CaI2 = 0.400 / 47.62 = 0.00840

7 0

3 years ago