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Marina CMI [18]
2 years ago
15

Jon does an experiment in which he varies the amount of calcium chloride (CaCl2) powder used for the chemical reaction with sodi

um bicarbonate (NaHCO3). He plans to observe whether the amount of CaCl2 powder has an effect on reaction rate.
Make the most logical prediction for the results of his experiment using the drop-down menus.

I predict that using a larger amount of calcium chloride powder will
______
a. increase
b. decrease
c. have no effect on

the reaction rate because the calcium chloride molecules will ______
a. have more kinetic energy
b. balance the number of product molecules
c. collide more often with other reactant molecules
Chemistry
1 answer:
asambeis [7]2 years ago
8 0

Answer:

a.) increase

c.) collide more often with other reactant molecules

Explanation:

Generally, when you increase the quantity of reactants, the rate of the reaction increases (the forward reaction). When there are more reactants, there are more chances for them to interact with each other. More interactions means more reactions.

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ra1l [238]
The simple equation used to calculate work is force multiplied by distance, thus as this is the case increasing the distance by a certain amount, assuming the force applied to the object is constant, the amount of work you are doing on the box for instance pushing it, is going to be greater

Since you are pushing the box with the same force covering a greater distance with the force.
4 0
2 years ago
In the chemical reaction:
tiny-mole [99]

Answer:

d

Explanation:

d

6 0
1 year ago
A ball sits on top of a hill. The wind blows and the ball begins rolling down the hill. What caused the ball to start moving?
ELEN [110]
The answer should be a
4 0
3 years ago
Read 2 more answers
g Consider a uniport system where a carrier protein transports an uncharged substance A across a cell membrane. Suppose that at
Alborosie

Answer :  The ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln Q\\\\\Delta G^o=-RT\times \ln (\frac{[A]_{inside}}{[A]_{outside}})

where,

\Delta G^o = standard Gibbs free energy  = -14.1 kJ/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 25^oC=273+25=298K

Q  = reaction quotient

[A]_{inside} = concentration inside the cell

[A]_{outside} = concentration outside the cell

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

-14.1\times 10^3J/mol =-(8.314J/K.mol)\times (298K)\times \ln (\frac{[A]_{inside}}{[A]_{outside}})

\frac{[A]_{inside}}{[A]_{outside}}=296.2

Thus, the ratio of the concentration of substance A inside the cell to the concentration outside is, 296.2

3 0
2 years ago
Suppose that 0.50 grams of barium-131 are administered orally to a patient. Approximately how many milligrams of the barium woul
nalin [4]

Two months later 13.8 milligrams of the barium-131 still be radioactive.

<h3>How is the decay rate of a radioactive substance expressed ? </h3>

It is expressed as:

A = A_{0} \times (\frac{1}{2})^{t/T}

where,

A = Amount remaining

A₀ = Initial Amount

t = time

T = Half life

Here

A₀ = 0.50g

t  = 2 months = 60 days

T = 11.6 days  

Now put the values in above expression we get

A = A_{0} \times (\frac{1}{2})^{t/T}

   = 0.50 \times (\frac{1}{2})^{60/11.6}

   = 0.50 \times (\frac{1}{2})^{5.17}

   = 0.50 × 0.0277

   = 0.0138 g

   = 13.8 mg          [1 mg = 1000 g]

Thus from the above conclusion we can say that Two months later 13.8 milligrams of the barium-131 still be radioactive.

Learn more about the Radioactive here: brainly.com/question/2320811

#SPJ1

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Suppose that 0.50 grams of ban that 0.50 grams of barium-131 are administered orally to a patient. Approximately many milligrams of the barium would still be radioactive two months later? The half-life of barium-131 is 11.6 days.

3 0
1 year ago
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