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ankoles [38]
2 years ago
8

Calculate the number of moles of NaOH contained in 250. mL of a 0.05M solution.

Chemistry
2 answers:
Katena32 [7]2 years ago
4 0

Answer:

.0125 moles

Explanation:

Molarity = moles/volume

.05 M = moles/.250L

.0125 = moles

topjm [15]2 years ago
3 0

Answer:

https://newshamchem.weebly.com/uploads/2/2/0/3/22035226/molarity_&_worksheet_answer.pdf

question 9.

Explanation:

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The number of protons in the nucleus of the atom is equal to the atomic number. True False
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Yes it is equal,True 
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Which has greater kinetic energy: a bullet that has a mass of 0.05kg travelling at 2,500m/s, or a lorry that has a mass of 4,500
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the answer will be lorry

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Determine the resulting pH when 12mL if 0.16M HCl are reacted with 32 mL if 0.24M KOH.
TEA [102]

Answer:

pH = 13.1

Explanation:

Hello there!

In this case, according to the given information, we can set up the following equation:

HCl+KOH\rightarrow KCl+H_2O

Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:

n_{HCl}=0.012L*0.16mol/L=0.00192mol\\\\n_{KOH}=0.032L*0.24mol/L=0.00768mol

Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:

n_{KOH}=0.00768mol-0.00192mol=0.00576mol

And the resulting concentration of KOH and OH ions as this is a strong base:

[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M

And the resulting pH is:

pH=14+log(0.131)\\\\pH=13.1

Regards!

3 0
3 years ago
How many moles of calcium nitrate would react with 4.55 moles of chromium(iii) sulfate to produce calcium sulfate and chromium(i
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The equation for the reaction between the calcium nitrate and chromium sulfate to form calcium sulfate and chromium(iii) nitrate is,
                            3Ca(NO3)2 + Cr2(SO4)3 --> 3CaSO4 + 2Cr(NO3)3
The balanced equation shows that for every 1 mole of chromium sulfate, there will be 3 moles of calcium nitrate. Therefore for this item, there will be 4.55 moles times 3 which is equal to 13.65. 
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A student dissolves 6.2g of aniline c6h5nh2 in 350.ml of a solvent with a density of 1.04/gml . the student notices that the vol
Troyanec [42]

Answer : The correct answer for molarity = 0.19 \frac{mol}{L} and Molality = 0.18 \frac{mol}{Kg}.

Given : Mass of aniline = 6.2 g

Volume of solvent = 350 mL Density of solvent = 1.04 g/mL

1) Molarity : It is defined as number of moles of solute present in Litre of solution . It is expressed as :

Molarity(M) = \frac{moles of solute(mole)}{volume of solution (L)}\

Molairty can be found in following steps :

Step 1 : To calculate mole :

Mole of solute can be calculate using mole formula :

Mole of solute = \frac{given mass of solute (g)}{molar mass of solute\frac{g}{1 mol}}

Molar mass of aniline (C₆H₅NH₂) = 93.13 \frac{g}{1 mol}

Mass of aniline = 6.2 g

Plugging values in mole formula :

Mole  =  \frac{6.2 g}{93.13 \frac{g}{1 mol}}

Mole = 0.0665 mol

Step 2 : To find volume of solution

Volume of solution = volume of solute + volume of solution

Since addition of aniline does not change final volume , so volume of solvent = volume of solution. Since volume is given in mL , so it need to be converted to L .

1 L = 1000mL

Volume of solution = \frac{350 mL}{1000mL}  * 1 L

Volume of solution = 0.350 L

Step 3: Plug value of mole and volume in molarity formula :

Molarity = \frac{0.0665 mol}{0.350 L }

Molarity = 0.19 M or 0.19 \frac{mol}{L}

------------------------------------------------------------------------------------------

2) Molality : It is defined as mole of solute present in Kilogram of solvent . It can be expressed as :

Molality (m) = \frac{mole of solute (mol)}{Kilogram of solvent (Kg)}

Following are the steps to calculate molality :

Step 1: To find mole of Solute :

Mole of solute can be found out using mole formula . It is same as done for molarity .

Mole = 0.0665 mol

Step 2 : To find kilogram of solvent :

Mass of solvent can be calculated using density formula as :

Density \frac{g}{mL} = \frac{mass (g) }{volume (mL)}

Plugging value in density formula :

1.04 \frac{g}{mL}  =  \frac{ mass }{350 mL}

Multiplying both side by 350 mL

1.04 \frac{g}{mL} * 350 mL = \frac{x}{350 mL } * 350 mL

Mass of solvent = 364 g

Since mass is in g, it need to be converted to Kg . ( 1 Kg = 1000 g )

Mass of solvent = \frac{364 g}{1000g} * 1 Kg

Mass of solvent = 0.364 Kg

Step 3: Plug values of mole and Kg in molality formula :

Molality = \frac{0.0665 mol}{0.364 Kg}

Molality = 0.18 m or 0.18 \frac{mol}{Kg}

3 0
3 years ago
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