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2 years ago

Please help

Chemistry

2 answers:

Scilla [17]2 years ago

5 0

B .metaphase is the answer

vitfil [10]2 years ago

4 0

The chromosomes line up towards the center during B. Metaphase

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For the reaction between aqueous solutions of acetic acid (CH3COOH) and barium hydroxide, Ba(OH)2, 1. Write the balanced molecul

Crazy boy [7]

Answer:

$2CH_3COOH(aq)+Ba(OH)_2(aq)\rightarrow Ba(CH_3COO)_2(aq)+2H_2O(l)$

Explanation:

When acetic acid solution and barium hydroxide solution react together to give an aqueous solution of barium acetate and water

The balanced chemical reaction will be given by

$2CH_3COOH(aq)+Ba(OH)_2(aq)\rightarrow Ba(CH_3COO)_2(aq)+2H_2O(l)$

8 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

2 years ago

143°C = _____ NEED HELP NOW!!!

Strike441 [17]

289.4 F is the right one man

5 0

3 years ago

Read 2 more answers

Law of Conservation of Mass in terms of atoms

VARVARA [1.3K]

No atoms are lost or made during the chemical reaction so the total mass of the products is equal to the total mass of the reactants. In an atom, protons and neutrons contribute to the mass and since the number of them doesn’t change, the mass doesn’t either.

4 0

2 years ago

Which of the following elements is rarely found in organic molecules?

fgiga [73]

Answer:

C: Carbon

Explanation:

Hydrogen, oxygen, and carbon are the most important elements in organic compounds.

8 0

3 years ago