All categories

3 years ago

How many grams of NaoH are needed to nutralize 50 grams of H2SO4?

1 answer:

6 0

to neutralize 1 mole of H2 S o4 we need one mole of any if we are having 50 grams of H2 S o4 it means the mole of H2 S o4 in 50 gram will be 50×40)/98 hence

utilising 50 grams of H2 S o4 we need approximately 20. 5 gm of Naoh

You might be interested in

Please help i have untill tomrow 2SO2(g) + O2(g) ----> 2SO3(g)

siniylev [52]

Answer: mmmmmm asking for mrs.howard work I see lol good luck grace

Explanation:

8 0

3 years ago

Tropical rain forests are being cut down to create farms and pastures, and to harvest trees for construction. What has this type

Mrrafil [7]

The deforestation of tropical rainforest trees to create farms and pastures, and to harvest trees for construction has increased soil erosion.

<h3>WHAT IS DEFORESTATION:</h3>

Deforestation is the process whereby trees are cut down for commercial purposes e.g. lumbering for paper production.

Rainforest is a vegetation characterized by trees and shrubs. However, these trees serve as soil cover and help prevent soil erosion.

This means that deforestation of tropical rainforest trees to create farms and pastures, and to harvest trees for construction has increased soil erosion

Learn more about deforestation at: brainly.com/question/11697527

5 0

2 years ago

*pls hurry, will give brainlyest or whatever!!*

brilliants [131]

Answer:

all of the above. they all are chemical reactions

4 0

3 years ago

a sample of fully saturated clay weighs 1350 g in its natural state and 975 g after drying. what is the natural water content of

svetoff [14.1K]

Answer:

The soil has a percentage of water by mass of 27.8 %.

Explanation:

Keeping in mind that

Mass of Clay = Mass of water + Mass of Dry Soil

we can <u>calculate the mass of water</u>:

1350 g = Mass of Water + 975 g

Mass of Water = 375 g

We can then <u>calculate the mass percentage of water in the soil</u>:

375 / 1350 * 100% = 27.8 %

7 0

3 years ago

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl ( aq ) , as describe

Nataly [62]

Answer:

0.9483 grams of manganese dioxide should be added to excess HCl.

Explanation:

Pressure of the chlorine gas = P = 795 Torr = 1.046 atm (1 atm = 760 Torr)

Volume of the chlorine gas = V = 255 ml = 0.255 L

Temperature of the chlorine gas = T = 25°C= 298.15 K

Moles of chlorine gas = n

Using ideal gas equation:

PV = nRT

$n=\frac{PV}{RT}=\frac{1.046 atm\times 0.255 L}{0.0821 atm l/mol k\times 298.15 K}$

n = 0.01090 mol

$MnO_2 ( s ) + 4 HCl ( aq)\rioghtarrow MnCl_2 ( aq ) + 2 H_2O ( l ) + Cl_2 ( g )$

According to reaction , 1 mole of chlorine gas is obtained from 1 mole of manganese dioxide.

Then 0.01090 moles of chlorine gas will be obtained from:

$\frac{1}{1}\times 0.01090 mol=0.01090 mol$ manganese dioxide

Mass of 0.01090 moles of manganese dioxide:

0.01090 mol × 87 g/mol = 0.9483 g

8 0

3 years ago

How many grams of NaoH are needed to nutralize 50 grams of H2SO4?​

How many grams of NaoH are needed to nutralize 50 grams of H2SO4?