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G To Ml

 Aug     


General search term.
Might be referring to: the conversion from grams to ml for water is extremely easy. One gram of pure water is exactly one milliliter..
This is a seasonal trend that repeats every August.
This trend is forecasted to be growing in the next 2 years.

  

Track this trend
(Monitor this trend over time)
 12 Months
Average: 37.56%
Trend: growing
MOM change: 0.07%
 5 Years
Average: 32.2%
Trend: growing
MOM change: 0.67%

Latest forum discussions

We tracked 3409 total in the last 3 months

RE:Stökiometri- Hur mycket väger 25 L svavelsyra om densiteten vid 293,25 K är 1,555 g/ml?
Jag försökte lösa denna genom att multiplicera 25 dm^3 och 1.555 g/ml och fick ungefär 39 kg vilket är rätt svar men undrar om detta var rätt sätt att lösa uppgiften på eller om jag bara hade tur. Förstår inte riktigt varför temperaturen nämndes om den inte ska användas för att lösa uppgiften.
Chazzy
Published on 2021-02-07 in the Kemi 1 forum
This discussion is in Swedish.
How to prepare 100 ml of 25% v/v glutaldehyde solution from 50% wt% glutaldehhyde in water?
Hi everyone. I want to prepare a 100 ml 25% v/v glutaldehyde solution and the glutaldehyde available is 50% wt% in water, MW= 100.12 g/mol, d= 1.106 g/ml. How would I go about doing it? Thanks.
Mphoqile Mokone
Published on 2021-02-05 in the Acetic Acid forum
This discussion is in English.
How to prepare 100 ml of 25% v/v glutaldehyde solution from 50% wt% glutaldehhyde in water?
Hi everyone. I want to prepare a 100 ml 25% v/v glutaldehyde solution and the glutaldehyde available is 50% wt% in water, MW= 100.12 g/mol, d= 1.106 g/ml. How would I go about doing it? Thanks.
Mphoqile Mokone
Published on 2021-02-05 in the Acetic Acid forum
This discussion is in English.
RE:d(g/ml) 인 염산수용액 만들 때
HCl ( 36.5 g/mL, 36.5%, D=1.179 g/mL) 밀도는 인터넷에서 36%일 경우 1.179 g/mL 였는데 시약병에 써있는 것을 사용하세요 ​ (1mol / 36.5 g/mL) x 0.365 x 1.179 g/mL x 1000 mL = 11.79 M 11.79 M x 10 mL = (A) x 1000 A = 0.1179 M ​ 0.1179 mol/L
아샤 님 답변
Published on 2020-12-06 in the 화학, 화학공학 forum
This discussion is in Korean.
RE:d(g/ml) 인 염산수용액 만들 때
HCl ( 36.5 g/mL, 36.5%, D=1.179 g/mL) 밀도는 인터넷에서 36%일 경우 1.179 g/mL 였는데 시약병에 써있는 것을 사용하세요 ​ (1mol / 36.5 g/mL) x 0.365 x 1.179 g/mL x 1000 mL = 11.79 M 11.79 M x 10 mL = (A) x 1000 A = 0.1179 M ​ 0.1179 mol/L
아샤 님 답변
Published on 2020-12-06 in the 화학, 화학공학 forum
This discussion is in Korean.
What is the molality of a solution consisting of 52.7 mL of benzene (C6H6; d = 0.877 g/mL) in 239 mL of hexane (C6H14; d = 0.660 g/mL)?
What is the molality of a solution consisting of 52.7 mL of benzene (C6H6; d = 0.877 g/mL) in 239 mL of hexane (C6H14; d = 0.660 g/mL)?
Anonymous
Published on 2021-02-04 in the Science & Mathematics forum
This discussion is in Unknown.
What is the molality of a solution consisting of 52.7 mL of benzene (C6H6; d = 0.877 g/mL) in 239 mL of hexane (C6H14; d = 0.660 g/mL)?
What is the molality of a solution consisting of 52.7 mL of benzene (C6H6; d = 0.877 g/mL) in 239 mL of hexane (C6H14; d = 0.660 g/mL)?
Anonymous
Published on 2021-02-04 in the Science & Mathematics forum
This discussion is in Unknown.
RE:How many milliliters of concentrated hydrochloric acid solution (36.0% HCl by mass, density = 1.18 g/mL) are required to produce 18.0 L ?
... mass, density = 1.18 g/mL) are required to produce 18...x volume                                                                                      = 1.18 g/mL x 1000 mL                                                                                      = 1180 gramsBut the ...molar mass                                         = 424.8 g/ 36.5 (g/mol)                                         = 11.64 ...
jacob s
Published on 2020-12-12 in the Science and Mathematics forum
This discussion is in English.
RE:How many milliliters of concentrated hydrochloric acid solution (36.0% HCl by mass, density = 1.18 g/mL) are required to produce 18.0 L ?
... mass, density = 1.18 g/mL) are required to produce 18...x volume                                                                                      = 1.18 g/mL x 1000 mL                                                                                      = 1180 gramsBut the ...molar mass                                         = 424.8 g/ 36.5 (g/mol)                                         = 11.64 ...
jacob s
Published on 2020-12-12 in the Science and Mathematics forum
This discussion is in English.
The density of an unknown liquid was calculated to be 0.78 g/ml. If the accepted value is 0.75 g/ml, calculate the percent error made.?
The density of an unknown liquid was calculated to be 0.78 g/ml. If the accepted value is 0.75 g/ml, calculate the percent error made.?
Iris
Published on 2021-01-29 in the Science and Mathematics forum
This discussion is in Unknown.
The density of an unknown liquid was calculated to be 0.78 g/ml. If the accepted value is 0.75 g/ml, calculate the percent error made.?
The density of an unknown liquid was calculated to be 0.78 g/ml. If the accepted value is 0.75 g/ml, calculate the percent error made.?
Iris
Published on 2021-01-29 in the Science and Mathematics forum
This discussion is in Unknown.
The density of an unknown liquid was calculated to be 0.78 g/ml. If the accepted value is 0.75 g/ml, calculate the percent error made.?
The density of an unknown liquid was calculated to be 0.78 g/ml. If the accepted value is 0.75 g/ml, calculate the percent error made.?
Iris
Published on 2021-01-29 in the Science and Mathematics forum
This discussion is in Unknown.
The density of an unknown liquid was calculated to be 0.78 g/ml. If the accepted value is 0.75 g/ml, calculate the percent error made.?
The density of an unknown liquid was calculated to be 0.78 g/ml. If the accepted value is 0.75 g/ml, calculate the percent error made.?
Iris
Published on 2021-01-29 in the Science and Mathematics forum
This discussion is in Unknown.
The density of an unknown liquid was calculated to be 0.78 g/ml. If the accepted value is 0.75 g/ml, calculate the percent error made.?
The density of an unknown liquid was calculated to be 0.78 g/ml. If the accepted value is 0.75 g/ml, calculate the percent error made.?
Iris
Published on 2021-01-29 in the Science and Mathematics forum
This discussion is in Unknown.
The density of an unknown liquid was calculated to be 0.78 g/ml. If the accepted value is 0.75 g/ml, calculate the percent error made.?
The density of an unknown liquid was calculated to be 0.78 g/ml. If the accepted value is 0.75 g/ml, calculate the percent error made.?
Iris
Published on 2021-01-29 in the Science and Mathematics forum
This discussion is in Unknown.
The density of solid sand (without air spaces) is about 2.84 g/mL. The density of gold is 19.3 g/mL.?
If a 1.00 kg sample of sand containing some gold has a density of 3.10 g/mL (without air spaces), what is the percentage of gold in the sample?
pantomath
Published on 2020-12-05 in the Science and Mathematics forum
This discussion is in English.
The density of solid sand (without air spaces) is about 2.84 g/mL. The density of gold is 19.3 g/mL.?
If a 1.00 kg sample of sand containing some gold has a density of 3.10 g/mL (without air spaces), what is the percentage of gold in the sample?
pantomath
Published on 2020-12-05 in the Science and Mathematics forum
This discussion is in English.
The density of solid sand (without air spaces) is about 2.84 g/mL. The density of gold is 19.3 g/mL.?
If a 1.00 kg sample of sand containing some gold has a density of 3.10 g/mL (without air spaces), what is the percentage of gold in the sample?
pantomath
Published on 2020-12-05 in the Science & Mathematics forum
This discussion is in English.
The density of solid sand (without air spaces) is about 2.84 g/mL. The density of gold is 19.3 g/mL.?
If a 1.00 kg sample of sand containing some gold has a density of 3.10 g/mL (without air spaces), what is the percentage of gold in the sample?
pantomath
Published on 2020-12-05 in the Science & Mathematics forum
This discussion is in English.