HELP PLEASE!!! time limited

Is ivy a monocot or dicot

Aleksandr-060686 [28]

Answer:

dicot

Explanation:

I looked it up lmoooo

7 0

3 years ago

What is the percent by volume of isopropyl alcohol in a solution made by mixing 120 mL of the alcohol with enough water to make

Elena L [17]

Answer: The percent by volume of isopropyl alcohol in a solution made by mixing 120 mL of the alcohol with enough water to make 350 mL of solution is 25.5%.

Explanation:

Given: Volume of solute = 120 mL

Volume of solvent = 350 mL

Now, total volume of the solution is as follows.

$V_{total} = V_{solute} + V_{solvent}\\= 120 mL + 350 mL\\= 470 mL$

Let us assume that 100 mL of solution is taken and the amount of isopropyl alcohol present in it is as follows.

$\frac{V_{solute}}{V_{total}} \times 100 mL\\\frac{120 mL}{470} \times 100 mL\\= 25.53 mL$

Hence, there is 25.53 mL isopropyl alcohol is present in 100 mL of solution. Therefore, %v/v is calculated as follows.

$Percent (v/v) = \frac{25.53 mL}{100 mL}\\= 25.5%$

Thus, we can conclude that the percent by volume of isopropyl alcohol in a solution made by mixing 120 mL of the alcohol with enough water to make 350 mL of solution is 25.5%.

7 0

3 years ago

Complete the chart. Atomic Number: 10. 1s

Goshia [24]

2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f
6s 6p 6d
7s 7p
hope it helps

7 0

3 years ago

A student was asked to determine the concentration of ammonia, a volatile substance used in the clinical setting as a respiretor

UNO [17]

This question is describing two chemical equations whereby the concentration of ammonia has to be determined. The first reaction is between 25.00 mL of ammonia and 50.00 mL of 0.100-M HCl whose excess was neutralized with 21.50 mL of 0.050-M Na₂CO₃ and thus, the concentration ammonia in the cloudy solution was determined as 0.114 M.

First of all we need to go over the titration of the excess HCl with Na₂CO₃ by writing the chemical equation it takes place when they react:

$2HCl+Na_2CO_3\rightarrow 2NaCl+CO_2+H_2O$

Whereas the mole ratio of HCl to Na₂CO₃ is 2:1 and the volume of the HCl leftover is determined as follows:

$V_{HCl}^{leftover}=\frac{2*0.050M*21.50mL}{0.100M} =21.5mL$

Next, we infer that the consumed volume of HCl by the ammonia solution was:

$V_{HCl}^{consumed}=50.00mL-21.50mL=28.5 mL$

Then, we write the chemical equation that takes place between ammonia and HCl:

$HCl+NH_3\rightarrow NH_4Cl$

Whereas the mole ratio is now 1:1, which means that the concentration of ammonia was:

$M_{NH_3}=\frac{28.5mL*0.100M}{25.00mL}\\\\ M_{NH_3}=0.114M$

Learn more:

(Titration) brainly.com/question/15687419
(Titration) brainly.com/question/25328286

8 0

2 years ago

Concentration required to begin precipitate pbcl2 for pbcl2 ksp=1. 17×10−5

Ne4ueva [31]

The concentration required to begin to precipitate PbCl2 for PbCl2 is 0.0216 M.

<h3>What is molarity?</h3>

Molarity is the measure of the concentration of any solute in per unit volume of the solution.

The reaction is $\rm Pb^2^+(aq) + 2Cl^-(aq) \rightarrow PbCl_2(s).$

The molarity of lead is 0.025 M

The ksp is given 17×10⁻⁵

Now, calculating the concentration

$[Pb^2^+] = 0.025 M.\\Ksp = 1.17 \times 10^-^5\\Ksp = [Pb^2^+] \times [Cl^-]^2\\[Cl^-] = \dfrac{\sqrt{ Ksp}}{[Pb^2^+]} \\\\[Cl^-] = \dfrac{\sqrt{ 0.0000117}}{0.025} \\[Cl^-] = 2.16 \times 10^-^2M.$

Thus, the concentration required to begin to precipitate PbCl2 for PbCl2 is 0.0216 M.

Learn more about molarity

brainly.com/question/10725862

#SPJ4

5 0

2 years ago