A doubling of the concentration of a doubles the rate of the reaction. true or false

write the balanced molecular chemical equation for the reaction in aqueous solution for sodium hydroxide and tin(iv) acetate. if

dimaraw [331]

The reaction of Sodium hydroxide and Tin(iv) acetate is as follows:

Sodium Hydroxide + Tin(IV) Acetate = Sodium Acetate + Tin(IV) Hydroxide

NaOH + Sn(C2H3O2)4 = C2H3NaO2 + Sn(OH)4

Balanced Chemical Equation:

4NaOH + Sn(C2H3O2)4 → 4C2H3NaO2 + Sn(OH)4

What is an Aqueous solution?

An aqueous solution is one in which water serves as the solvent. It is also known as a water-based solution. Many common chemical substances are dissolved in water, and these are called aqueous solutions. Examples of aqueous solutions include saltwater, vinegar, and sugar water.

What is Sodium hydroxide?

Sodium hydroxide, also known as lye or caustic soda, is a white, solid crystalline substance used in a number of industrial and household applications. It is a strong base that is highly soluble in water and can be used to adjust the pH of a solution. It is also used in the manufacture of soaps, detergents, and cleaning products.

What is Tin(IV)?

Tin(IV) is the chemical name for tin, which is a chemical element with the atomic number of 50. It is a silvery-white, malleable metal that is used in a variety of applications, including alloys, solders, and some electronics.

To know more about Sodium hydroxide,

LINK- brainly.com/question/29509565

CODE- #SPJ4

5 0

8 months ago

Nitrogen has an atomic number of 7; therefore, it has ______ electrons in its outermost electron shell.

galina1969 [7]

it has 5 electrons in it's outermost shell.
s 2e- , p 5e-

4 0

3 years ago

Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

8 0

3 years ago

A little aluminum boat (mass of 14.50

Valentin [98]

When the volume is 450 cm3 so the boat will displace 450 g.

so we have 450 g - 14.5g = 435.5 left

so to get how many pennies can be added to the boat before it sinks can be determined by:

divided the mass for the one penny:

no.of Penny = 435.5 / 2.5g

= 174.2

∴ the boat will float with 174 pennies and will start to sink with 175

5 0

2 years ago

Can someone help? Name the compound:

Serhud [2]

CH=benzene

Why?

benzene is represented by the empirical formula CH, which indicates that a typical sample of the compound contains one atom of carbon (C) to one atom of hydrogen (H).

----(Is this what you meant???)

5 0

2 years ago