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pochemuha
7 days ago
9

A 1.0 mole pure sample of molten tin is dissolved in a 5.0 mole pure sample of molten copper. The solution is set aside to cool

and solidify. The atomic radius of tin is 140 pm and the atomic radius of copper is 128 pm. a. Identify the type of alloy that is formed. Justify your answer. b. Identify the solvent in this solution. Justify your answer. ​
Chemistry
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Los automóviles actuales tienen “parachoques de 5 mi/h (8 km/h)” diseñados para comprimirse y rebotar elásticamente sin ningún d
KiRa [2933]

Response: k = 23045 N/m

Clarification:

To determine the spring constant, one must consider the maximum elastic potential energy that the spring can withstand. The kinetic energy of the vehicle should equal at minimum the elastic potential energy of the spring when it is fully compressed. Hence, we express it as:

K=U\\\\\frac{1}{2}Mv^2=\frac{1}{2}kx^2    (1)

M: mass of the vehicle = 1050 kg

k: spring constant =?

v: car speed = 8 km/h

x: maximum spring compression = 1.5 cm = 0.015m

You need to resolve equation (1) for k. Beforehand, convert the speed v to meters per second:

v=8\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=2.222\frac{m}{s}

k=\frac{Mv^2}{x^2}=\frac{(1050kg)(2.222m/s)^2}{(0.015m)^2}=23045\frac{N}{m}

The spring constant calculates to 23045 N/m

3 0
2 months ago
Some fruits and vegetables are preserved by pickling them. Nandini got confused
Anarel [2989]

Answer:

yes

Explanation:

yes

3 0
3 months ago
First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta
Alekssandra [3086]

The rate equation for a first order reaction can be expressed as follows:

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}

In this context, k represents the reaction's rate constant, t denotes the time the reaction takes, A_{0} is the initial concentration, and A_{t} is the concentration at time t.

The rate constant of the reaction is 0.1 day^{-1}.

(a) If we start with an initial concentration of 100, when 90% of the substance is eliminated, the remaining quantity at time t will be 100-90=10. By substituting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days

The time required to destroy 90% of the substance amounts to 23.03 days.

(b) If the initial concentration is set at 100, when 99% is destroyed, the present amount at time t will be 100-99=1. By substituting the input values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days

This results in a duration of 46.06 days required to eradicate 99% of the chemical.

(c) Should the initial concentration be set at 100, with 99.9% of the chemical removed, the remaining quantity at time t will be 100-99.9=0.1. Substituting the values yields

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days

Thus, the time needed to eliminate 99.9% of the chemical is calculated as 69.09 days.

5 0
2 months ago
A student is given two 10g samples, each a mixture of only NaCl(s) and KCl(s) but in different proportions. Which of the followi
alisha [2963]

Answer:

To figure out which sample contains more KCl, information on the mass of chlorine in each mixture is needed; this can be found using the law of definite proportions.

Explanation:

The law of definite proportions explains that a chemical compound always contains the same proportion by mass of its elements. For example, a compound made of two elements maintains consistent ratios regardless of sample size.

By applying this law, one can calculate the ratios of sodium to chlorine in NaCl and potassium to chlorine in KCl if the chlorine mass is known. This lets us identify which mixture holds a larger percentage of KCl.

6 0
3 months ago
Read 2 more answers
Calculate the equilibrium constant, k, for the reaction shown at 25 °c. fe3 (aq) b(s) 6h2o(l)⟶fe(s) h3bo3(s) 3h3o (aq) the balan
lions [2927]

The complete question is:

Determine the equilibrium constant, K, for the given reaction at 25 °C.

Fe3+(aq)+B(s)+6H2O(l)-------->Fe(s)+H3BO3(s)+3H3O+(aq)

The balanced reduction half-reactions for this equation and their corresponding standard reduction potential values (E°) are

Fe3+(aq)+3e------>Fe(s) E°=-0.04V

H3BO3(s)+3H3O+(aq)+3e------->B(s)+6H2O(l) E°=0.8698 V

Answer:

1.05

Explanation:

E° cell= E°cat - E°an

E°cat = 0.8698 V

E°an= -0.04V

E°cell=0.8698 V - (-0.04V)

E°cell= 0.9098 V

E°cell= 0.0592/n logK

From the balanced reduction reaction equations, n=3

0.9098 V= 0.0592/3 logK

logK= 0.0217

K= Antilog (0.0217)

K= 1.05

3 0
2 months ago
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