When would sediment get deposited in a stream

Which of the following is the correct way to express 0.0000321 in scientific notation?

scZoUnD [109]

The answer is 3.21*10^-5

7 0

3 years ago

If the partial pressure of a gas produced in a reaction was determined to be 0.35 atm in a 161.0 ml container and at a temperatu

nirvana33 [79]

Thank you for posting your question here at brainly. I hope the answer will help you. The number of moles of gas were produced the answer will be 0.0023 mol.

Feel free to post your question here at brainly, if you have more questions.

7 0

3 years ago

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Which statement is true because of newtons second law ?

katrin2010 [14]

Newton's second law, by definition, states that, The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. Therefore, based on the given options above, the correct answer would be option D: when the net force on an object decreases, the objects acceleration decreases.

7 0

3 years ago

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Bleach contains the active ingredient NaClO. Analysis of bleach involves two sequential redox reactions: First, bleach is reacte

Alisiya [41]

Answer:

0,31%

Explanation:

For the reaction:

I₂ + 2 S₂O₃²⁻ → 2 I⁻ + S₄O₆²⁻

0,043 L × 0,117 M of sodium tiosulfate = 5,031x10⁻³ moles of S₂O₃²⁻

5,031x10⁻³ moles of S₂O₃²⁻ × $\frac{1 I_2 mol}{2 S_{2}O_3 mol^{2-}}$ = 2,5156x10⁻³ moles of I₂

These moles of I₂ were produced from:

ClO⁻⁻ + 2 H⁺ + 2 I⁻ → I₂ + Cl⁻ + H2O

2,5156x10⁻³ moles of I₂ ≡ moles of NaClO

2,5156x10⁻³ moles of NaClO × $\frac{74,44 g}{1mol}$ = 0,187 g of NaClO

Thus, percentage composition by mass is:

$\frac{0,187 g of NaClO}{60 g Of Bleach} x 100$ = 0,31%

I hope it helps!

5 0

4 years ago

How much energy (heat) is required to convert 248 g of water from 0oC to 154oC? Assume that the water begins as a liquid, that t

Arturiano [62]

Answer: The amount of heat required is 775.7 kJ

Explanation:

The processes involved in the given problem are:

$1.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\2.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\3.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\4.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(154^oC,427K)$

Now, we calculate the amount of heat released or absorbed in all the processes.

For process 1:

$q_1=m\times L_f$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 248 g

$L_f$ = latent heat of fusion = 334 J/g

Putting all the values in above equation, we get:

$q_1=248g\times 334J/g=84832J$

For process 2:

$q_2=m\times C_{p,l}\times (T_{2}-T_{1})$

where,

$q_2$ = amount of heat absorbed = ?

$C_{p,l}$ = specific heat of water = 4.184 J/g°C

m = mass of water = 248 g

$T_2$ = final temperature = $100^oC$

$T_1$ = initial temperature = $0^oC$

Putting all the values in above equation, we get:

$q_2=248g\times 4.184J/g^oC\times (100-0)^oC=103763.2J$

For process 3:

$q_3=m\times L_v$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of water or ice = 248 g

$L_v$ = latent heat of vaporization = $40.79kJ/mol\times \frac{1000}{18}=2266.1J/g$ (Conversion factor used: 1 kJ = 1000 J and molar mass of water = 18 g/mol)