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Anestetic [448]

2 years ago

13

When an Alka-Seltzer® tablet is put in water, there is a chemical change because there are bubbles produced. The Alka-Seltzer® t

ablet slowly disappears over time. What would speed up the chemical change?
Adding more water would cause the chemical change to speed up.
Changing the color of the tablet would cause the chemical change to speed up.
Decreasing the temperature of the water would cause the chemical change to speed up.
Increasing the temperature of the water would cause the chemical change to speed up.

2 answers:

egoroff_w [7]2 years ago

4 0

It’s either increasing the temp or water. the temp increasing adds more energy bc heat = more energy. so i would go w increasing temp

Soloha48 [4]2 years ago

3 0

Answer:

Heyy

Explanation:

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Calculate the number of atoms in 2 mol of sulfur dioxide (SO2).

Alex17521 [72]

Two moles of SO2 means 1.204x10^24 molecules, since there are 3 atoms in one molecule, multiply 1.204 x10^24 by 3 and you get 3.612x10^24.

7 0

3 years ago

How many water molecules does it contain? The density of water is 1.0 g/cm3.

max2010maxim [7]

Complete question is;

A drop of water has a volume of approximately 7 × 10⁻² ml. How many water molecules does it contain? The density of water is 1.0 g/cm³.

This question will require us to first find the number of moles and then use avogadro's number to get the number of water molecules.

<em><u>Number of water molecules = 2.34 × 10²¹ molecules</u></em>

We are given;

Volume of water; V = 7 × 10⁻² ml

Density of water; ρ = 1 g/cm³ = 1 g/ml

Formula for mass is; m = ρV

m = 1 × 7 × 10⁻²

m = 7 × 10⁻² g

from online calculation, molar mass of water = 18.01 g/mol

Number of moles(n) = mass/molar mass

Thus;

n = (7 × 10⁻²)/18.01

n = 3.887 × 10⁻³ mol

from avogadro's number, we know that;

1 mol = 6.022 × 10²³ molecules

Thus,3.887 × 10⁻³ mol will give; 6.022 × 10²³ × 3.887 × 10⁻³ = 2.34 × 10²¹ molecules

Read more at; brainly.in/question/17990661

6 0

3 years ago

DONT SKIP OR ELSE what solids, liquids, and gases does the human body produce

Marysya12 [62]

farting pooping peeing :3

6 0

3 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m

Bond [772]

Answer:

$m_{H_2O}=4.86gH_2O$

Explanation:

Hello,

In this case, the described chemical reaction is:

$C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O$

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

$n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6$

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

$n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6$

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

$m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O$

Best regards.

3 0

3 years ago

Determine the value of the equilibrium constant, KgoalKgoalK_goal, for the reaction CO2(g)⇌C(s)+O2(g)CO2(g)⇌C(s)+O2(g), Kgoal=?

NISA [10]

Answer:

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .

Explanation:

$CO_2(g)\rightleftharpoons C(s)+O_2(g)$

$K_{goal}=?$

$K_{goal}=\frac{[C][O_2]}{[CO_2]}$

$2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2(g)$ ..[1]

$K_1=\frac{[CH_3COOH][O_2]^2}{[CO_2]^2[H_2O]^2}$

$2H_2(g)+O2(g)\rightleftharpoons 2H_2O(l)$ ..[2]

$K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}$

$CH_3COOH(l)\rightleftharpoons 2C(s)+2H_2(g)+O_2(g)$ ..[3]

$K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}$

[1] + [2] + [3]

$2CO_2(g)\rightleftharpoons 2C(s)+2O_2(g)$

( on adding the equilibrium constant will get multiplied with each other)

$K=K_1\times K_2\times K_3$

$K=5.40\times 10^{-16}\times 1.06\times 10^{10}\times 2.68\times 10^{-9}$

$K=1.53\times 10^{-14}$

$K=\frac{[C]^2[O_2]^2}{[CO_2]^2}$

On comparing the K and $K_{goal}$ :

$K^2=K_{goal}$

$K_{goal}=\sqrt{K}=\sqrt{1.53\times 10^{-14}}=1.24\times 10^{-7}$

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .

4 0

3 years ago