How is density found

What does the law of conservation of energy imply?

zheka24 [161]

Answer:

B. That energy can only be converted between different forms

Explanation:

The law of conservation of energy implies that energy can only be converted between different forms but can neither be created nor destroyed.

Energy is the ability to do work. It is very important physical quantity that is essential to all life forms.

Energy according to the law of conservation of energy is not produced neither is it destroyed.

Energy is simply transformed from one form to another.

4 0

3 years ago

How does aspirin interact with other drugs

Damm [24]

The most common drugs that aspirin may interact with are: Anti-inflammatory painkillers, such as diclofenac, ibuprofen, indomethacin, and naproxen. Taken with aspirin, these can increase the risk of bleeding. Warfarin, an anticoagulant drug, or a blood thinner, which stops the blood from clotting.

7 0

3 years ago

Do CO2 and h2o have the same geometry

Svetach [21]

No, they do not. Carbon dioxide has a linear geometry because the lone pair and bond pair repulsion cancels out; however, water has a bent structure because only the oxygen atom possesses a lone pair which brings the bonding electron pairs closer.

7 0

3 years ago

What is AgNO3 + KCI-KNO3+AgCI. What type of reaction does this represent

PIT_PIT [208]

C4H10 + 6.5 O2 ----> 4CO2 + 5H2O

2C4H10 + 13 O2 ----> 8CO2 + 10H2O

1. Count the C on the left (4), put a 4 where the C on the right.

2. Count the H on the left (1), you have two on the right, so you multimply this two by 5. Put the 5 in front of the H2O

3. Count the O on the right. You have 4*2 + 5 = 13. You have two on the left, so you need 6.5 on the left.

4. Now multiply everything on the equation by two so you have nice integer numbers.

5. check you have the same amount of everything on each side.
Example C: left 8, right 8, etc.
I hope this helps. :)

6 0

3 years ago

A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction

Molodets [167]

Answer: The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For NaOH:

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol$

For sulfuric acid:

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, $2.925\times 10^{-3}$ moles of KOH will react with = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = $(4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, $2.925\times 10^{-3}$ moles of KOH will produce = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sodium sulfate

For sodium sulfate:

Moles of sodium sulfate = $1.462\times 10^{-3}moles$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M$

For sulfuric acid:

Moles of excess sulfuric acid = $3.498\times 10^{-3}mol$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M$

For NaOH:

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0}{0.050}=0M$

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

8 0

3 years ago