All categories

3 years ago

Need Help Pls ! Homework Sheet

Chemistry

1 answer:

My name is Ann [436]3 years ago

8 0

20 balanced
21 unbalanced
22 balanced
23 balanced
24 unbalanced
25 unbalanced

You might be interested in

A shopkeeper has a few bottles of soft drinks in his shop.But,these are not labelled.He has to serve the drinks on the demand on

zepelin [54]

Answer:

He will decide which drink is to be served to whom, by the use of litmus paper.

Explanation:

The litmus paper is the most common indicator to determine the acidity or basicity of a solution. Blue litmus paper changes its color to red when a solution changes from basic to acidic while red litmus paper changes its color to blue when the opposite occurs (acid → basic).

First of all the litmus paper strip, pH indicator, is immersed in a solution and allowed to pass between 10 and 15 seconds while keeping the strip submerged. Afterwards it is removed, and then the strip compares the color. If the color is diffuse, there is a color scale where it is determined which solution has alkaline or acidic pH

5 0

3 years ago

How many grams of N2 are in 44.8L at STP?

lutik1710 [3]

Answer: 10 i think

Explanation:

3 0

3 years ago

How many ug of nickel (Ni) are required to make 25.00 nanoliters of a 1.25 mol/L solution? Be sure to report your answer to the

devlian [24]

volume of Ni = 25 nL = 25 x 10⁻⁹ L

mol Ni = 25 x 10⁻⁹ L x 1.25 mol/L = 3.125 x 10⁻⁸

mass = mol x Ar Ni

mass = 3.125 x 10⁻⁸ x 59 g/mol

mass = 1.84 x 10⁻⁶ g = 1.84 μg

4 0

2 years ago

Read 2 more answers

The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the s

raketka [301]

Answer : The equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

Explanation :

The dissociation of acid reaction is:

$C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-$

Initial conc. c 0 0

At eqm. c-x x x

Given:

c = $7.0\times 10^{-2}M$

$K_a=6.3\times 10^{-5}$

The expression of dissociation constant of acid is:

$K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}$

$K_a=\frac{(x)\times (x)}{(c-x)}$

Now put all the given values in this expression, we get:

$6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}$

$x=2.1\times 10^{-3}M$

Thus, the equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

4 0

3 years ago

"solid potassium iodide decomposes into iodine gas and solid potassium. Write a a balanced chemical equation for this reaction"

wel

$KI (s) ----\ \textgreater \ K(s) + I_2 (g)$

Now, balance the equation:

$2KI (s) ----\ \textgreater \ 2K(s) + I_2 (g)$

$I_2$ in gaseous state exist as a diatomic molecule.

3 0

3 years ago