If 1 mile = 5,280ft. how many miles are in 11,245ft

A first order reaction has a rate

EleoNora [17]

Answer:

one-half

Explanation:

cuz for a first order reaction is a half life independent of concentration and constant over time

4 0

3 years ago

when propene gas (C3H6) is added to bromine liquid, another liquid is produced which has the formula C3H6Br2. what is the chemic

klio [65]

Answer:

C3H6 + Br2 → C3H6Br2

Explanation:

The reaction in which C3H6Br2 (1,2-Dibromopropane) is created is:

C3H6 + Br2 → C3H6Br2

We can see that the only difference between the product (C3H6Br2) and the known reactant (C3H6) of the reaction is two bromine atoms (Br2). Br2 is diatomic bromine - a molecule we get after combining two bromine atoms. This compound is a red-brown liquid at room temperature, which means that that is the liquid described in your question.

4 0

3 years ago

For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ

Komok [63]

Answer:

Check the explanation

Explanation:

Answer – Given, $H_3PO_4$ acid and there are three Ka values

$K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}$

The transformation of $H_2PO_4- (aq) to HPO_4^2-(aq)$ is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

= 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

= 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

= -log 6.2x10-8

= 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

= 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

6 0

3 years ago

For each of the esters provided, identify the alcohol and the carboxylic acid that reacted.

Veronika [31]

Answer:

52. The alcohol USED => methanol, CH3OH

The carboxylic acid USED => propanoic acid, CH3CH2COOH.

53. The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

Explanation:

52. To obtain Methyl propanoate, CH3CH2COOCH3, we simply react propanoic, CH3CH2COOH and methanol, CH3OH together as shown below:

CH3CH2COOH + CH3OH —> CH3CH2COOCH3 + H2O

The alcohol used: methanol, CH3OH

The carboxylic acid used: propanoic acid, CH3CH2COOH.

53. To obtain Ethyl methanoate, HCOOCH2CH3, we simply react

Formic acid, HCOOH and ethanol, CH3CH2OH together as show below:

HCOOH + CH3CH2OH —> HCOOCH2CH3 + H2O

The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

7 0

3 years ago

Solve and show work. Li2S + 2 HNO3 --> 2 LiNO3 + H2S (a) Calculate the mass of lithium sulfide that will react with 250 mL of

Elenna [48]

Li2S + 2 HNO3 --> 2 LiNO3 + H2S

Li2 S + H2 N2 O2 --> Li2 N2 O5 + H2 S

Li S + H2 N2 O5 -> Li N2 O5 + H2 S

Li2 S2 + H4 N4 O10 --> Li2 N4 O10 + H4 S2

Li^2 S^2 + H^4 N^4 O^10 --> Li^2 N^4 O^10 + H^4 S^2

7 0

3 years ago

If 1 mile = 5,280ft. how many miles are in 11,245ft​

If 1 mile = 5,280ft. how many miles are in 11,245ft