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3 years ago

1. How can you make a solid solute dissolve into solution faster?

1 answer:

3 0

The correct answer to your question is B,, make the solute particles smaller.
let me know if you have any further questions
:)

You might be interested in

Determine the number of atoms in 41.0 grams of calcium, Ca. (The mass of one mole of calcium is 40.08 g.)

Dovator [93]

Hi there!

Let me know if you have questions about my answer:

6.15 × 10²⁴

Explanation:

To find the number of atoms, divide the mass of calcium (m) by its molar mass (MM). This will give you the number of moles of calcium (n).

$n = m/MM$ Start with a formula

$=\frac{41.0g}{40.08g/mol}$ Substitute values from the question

$= \frac{41.0g*1mol}{40.08g}$ Simplify

$= \frac{41.0mol}{40.08}$ Cancel out the units "g"

$= 1.022...mol$ Number of moles of calcium, with 4 significant figures

Then, multiply the number of moles of calcium (n) by Avogadro's number ( $N_{A}$ ) to find the number of calcium atoms.

$Number.of.Ca=n*N_{A}$ Start with a formula

$= 1.022mol*\frac{6.022*10^{23}Ca}{1mol}$ Substitute values from the question

$= \frac{1.022mol*6.022*10^{23}Ca}{1mol}$ Simplify

$= 1.022*(6.022*10^{23}Ca)$ Cancel out the units "mol"

$= 6.154*10^{24} Ca$ Multiply, leave one extra digit to help round. "4" tells you to round down.

$= 6.15*10^{24} Ca$ Final answer to 3 significant figures*

*When multiplying and dividing numbers, your final answer will have the same number of significant figures as the number with the least number of significant figures in the question.

40.08 has 4 significant figures. 41.0 has 3 significant figures. So, you want 3 significant figures in your final answer.

I hope this helps!

Learn more about Avogadro's number here:

brainly.com/question/1445383

Check out another answer that uses significant figures:

brainly.com/question/3721164

7 0

2 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Part (a) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

Part (b) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

3 years ago

Help please first person to get it right gets my old Netflix account

spin [16.1K]

Butterflies are cold-blooded and need the light from the sun to warm the muscles they use to fly. Not only do butterflies like the sun, the plants the they thrive on need full direct sun. Most plants need at least 8 hours of sunlight to bloom properly and provide enough nectar.

7 0

2 years ago

Read 2 more answers

What is the pressure of a sample of gas at a volume of .335 L if it occupies 1700 mL at 850 mm Hg?

dangina [55]

Answer:

Pressure = 4313.43mmHg

Explanation:

P1 = ?

V1 = 0.335L

V2 = 1700mL =1700*10^-3L = 1.7L

P2 = 850mmhg

From Boyle's law, the volume of a fixed mass of gas is inversely proportional to its pressure provided that temperature remains constant.

P = k / v

K = pv. P1V1 = P2V2 = P3V3 =........=PnVn

P1V1 = P2V2

Solve for P1,

P1 = (P2*V2) / V1

P1 = (850 * 1.7) / 0.335

P1 = 4313.43mmHg

The pressure of the gas was 4313.43mmHg

7 0

3 years ago

Which of these statements is false? (A) Acetylene (ethyne) can be converted to the acetylide anion by treating with a strong bas

Free_Kalibri [48]

Answer:

Explanation:

Acetylene (C₂H₂) can be converted to the acetylide anion (C₂⁻²) when treated with a base because it will donate protons (2H⁺). So it will be a neutralization reaction. NaNH₂ and NaOH are strong bases because they are good electrons donators ( NaNH₂ has pair of electrons on N, and NaOH has the group OH⁻), but CH₃Li has no pair of electrons to donate, so it's not a strong base.

4 0

3 years ago