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3 years ago

Anton van Leeuwenhoek used a microscope to observe

Chemistry

1 answer:

Goshia [24]3 years ago

4 0

B. He was the first person to observe and identify living cells.

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an aqeous solution of oxalic acid h2c2o4 was prepared by dissolving a 0.5842g of solute in enough water to make a 100 ml solutio

vampirchik [111]

The question is incomplete, here is the complete question:

An aqeous solution of oxalic acid was prepared by dissolving a 0.5842 g of solute in enough water to make a 100 ml solution a 10 ml aliquot of this solution was then transferred to a volumetric flask and diluted to a final volume of 250 ml. How many grams of oxalic acid are in 100. mL of the final solution?

<u>Answer:</u> The mass of oxalic acid in final solution is 0.0234 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ ......(1)

Given mass of oxalic acid = 0.5842 g

Molar mass of oxalic acid = 90 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Molarity of oxalic acid solution}=\frac{0.5842\times 1000}{90\times 100}\\\\\text{Molarity of oxalic acid solution}=0.0649M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated oxalic acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted oxalic acid solution

We are given:

$M_1=0.0649M\\V_1=10mL\\M_2=?M\\V_2=250.0mL$

Putting values in above equation, we get:

$0.0649\times 10=M_2\times 250.0\\\\M_2=\frac{0.0649\times 10}{250}=0.0026M$

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of oxalic acid solution = 0.0026 M

Molar mass of oxalic acid = 90 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0026=\frac{\text{Mass of oxalic acid solution}\times 1000}{90\times 100}\\\\\text{Mass of oxalic acid solution}=\frac{0.0026\times 90\times 100}{1000}=0.0234g$

Hence, the mass of oxalic acid in final solution is 0.0234 grams

6 0

4 years ago

Please I need the answer asap

Nadusha1986 [10]

The energy needed to raise the temperature of water from 22.0ºC to 90.0ºC is c. 28.4 kJ.

<h3>What is specific heat?</h3>

The amount of energy needed to raise the temperature of one gram of a substance by one degree Celsius.

By the formula $Q = mc\Delta T$

Q is the heat

m is the mass

c is the specific heat

Now, c = 4.184 J/g.K

The change in temperature is 22.0 ºC to 90.0 ºC

Putting the value in the equation

$Q = 100 \times 4.184 \times (90.0 - 22.0)\\\\Q = 4.184 \times 68.0\\ \\Q= 284.5 \;J$

Thus, the energy needed to raise the temperature of water from 22.0ºC to 90.0ºC is 28.4 kJ

Learn more about specific heat

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4 0

2 years ago

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

Now put all the given values in the above formula, we get:

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

3 years ago

Can a mixture be formed into a compound?

trapecia [35]

No, i don’t think so

6 0

4 years ago

3NO2− + 8H+ + Cr2O72− → 3NO3− +2Cr3+ + 4H2O

Reptile [31]

According to this reaction:
3NO2^- + 8 H^+ + Cr2O7^-2 → 3NO3^- + 2Cr^+3 + 4 H2O
(1) 3 NO2^- → 3NO3^-, The oxidation state of nitrogen converted from +3 to +5
i.e oxidation process, ∴ NO2^- is reducing agent.
(2) Cr2O7^-2 → 2Cr^+3, The oxidation state of Cr converted from +6 to +3
i.e reduction process ∴ Cr2O7^-2 is an Oxidizing agent.
(3) H^+ → H2O, the oxidation state of Hydrogen is still +1, i.e No change (neither)

6 0

4 years ago