I need to know the answer to that

Which indicator is yellow in a solution with a pH of 9.8?

Vlada [557]

It's the Methyl Orange.
at about 4.4 pH, it changes from red to Yellow, to indicate an acid solution.
This pH indicator is normally used in titration of acids.

Hope this Helps :)

8 0

3 years ago

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If the initial volume is 55ml and the final volume after some rocks of sandstone are added is 65ml what is the volume of the chi

AfilCa [17]

Answer:

Volume = 10ml

Density = 1/5 g/ml or 0.20g/ml

Explanation:

The rocks are 10ml since the initial volume went up by 10.

Since density = mass/volume, you divide 2 by 10.

D = 2/10

D = 1/5 g/ml or 0.20g/ml

(Unit is g/ml aka grams/millileter)

7 0

3 years ago

Answer Quick

kow [346]

Answer:

20.9%

Explanation:

The percentage by mass of solution is given by dividing the mass of solute in grams by the mass of solution in grams then multiplying it by 100%.

% Mass of solution = mass of solute/mass of solution × 100%

= (27.0 g/ 129.0 g) × 100%

= 20.93%

= 20.9%

7 0

3 years ago

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The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10

LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

(4) The number of moles of $CO_2$ produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

Part 1 :

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

and,

As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

Part 2 :

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

Part 3 :

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

Part 4 :

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

and,

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

From this we conclude that,

As, 1 mole of $C_4H_{10}$ react to produce 4 moles of $CO_2$

As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

5 0

3 years ago

How many grams of sodium chloride are contained in 250.0 g of a 15% NaCl solution?

kramer

Given concentration of NaCl=15%

Means ,

In every 100g of Solution 15g of NaCl is present .

Now

Given mass=250g

So ,

$\\ \Large\sf\longmapsto 250\times 15\%$

$\\ \Large\sf\longmapsto 250\times \dfrac{15}{100}$

$\\ \Large\sf\longmapsto 37.5g$

37.5g of NaCl present in 250g of solution.

8 0

3 years ago

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