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2 years ago

Perform the following problem. Round your answer appropriately. 8.1456 - 5.12 =

Chemistry

1 answer:

il63 [147K]2 years ago

3 0

The difference between 8.1456 - 5.12 will be 3.0256 and rounding it off will give 3.

<h3>How to subtract the numbers?</h3>

It should be noted that the information given is that we should subtract 5.12 from 8.1456.

It should be noted that subtraction is used when we want to deduct a number from another number. It is taking away a number form another one.

In this case, we will take away 5.12 from 8.1456. This will give the difference between the numbers.

Therefore, 8.1456 - 5.12 = 3.0256

Therefore, the difference is 3.0256 and rounding it appropriately will give a value of 3.

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7 f Find the volume in dm3 and in mole of 0.505m of NaoH required to react with 40ml of 0.505m

Anna [14]

The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ $(n_{A}) = 1$

Mole ratio of the base, NaOH $(n_{B}) = 2$

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH $(M_{B}) = 0.505 M$

Volume of the acid, H₂SO₄ $(V_{A}) = 40 mL$

Molarity of the acid, H₂SO₄ $(M_{A}) = 0.505 M$

<h3>Volume of the base, NaOH $(V_{B})$ =? </h3>

$\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}$

Cross multiply

$0.505 * V_{B} = 20.2 * 2\\0.505 * V_{B} = 40.4$

Divide both side by 0.505

$V_{B} = \frac{40.4}{0.505}\\\\V_{B} = 80 mL$

Finally, we shall convert 80 mL to dm³. This can be obtained as follow:

$1000 mL = 1 dm^{3}\\\\Therefore,\\\\80 mL = \frac{80 mL * 1dm^{3}}{1000 mL}\\\\80 mL = 0.08dm^{3}$

Therefore, the volume of NaOH required is 0.08 dm³

Learn more: brainly.com/question/19053582

3 0

3 years ago

Which statement about the United Nations is an

Vinil7 [7]

The first one because some people might not believe that and others might believe that.

5 0

3 years ago

The value of the entropy change for the process N₂ (g) + 3H₂ (g) --> 2NH₃ (g) is ________.

kodGreya [7K]

Answer:

negative

Explanation:

Entropy is a measure of the "disorder" in a system.

In this reaction, the amount of disorder decreases. This is because one gas molecule (NH₃) has more order than two gas molecules (N₂ and H₂). Therefore, the entropy change should be negative.

5 0

2 years ago

The pH of a solution is 8. Some hydrochloric acid is added to the solution. Suggest the pH of the solution after mixing. pH=....

Molodets [167]

Answer:

Probably around 6 because the ph of hydrochloric acid is 3

Explanation:

3 0

3 years ago

Using the following equation 5KNO2+2KMnO4+3H2SO4=5KNO3+2MnSO4+K2SO4+3H20. Starting with 2.47 grams of KNO2 and excess KMnO4 how

Aleonysh [2.5K]

Given equation : $5KNO_{2} + 2KMnO_{4} + 3H_{2}SO_{4}\rightarrow 5KNO_{3} + 2MnSO_{4} + K_{2}SO_{4} + 3H_{2}O$

Given information = 2.47 grams KNO2 and excess KMnO4 and we need to find grams of water (H2O).

Since KMnO4 is in excess, so grams of water(H2O) can be calculated using grams of KNO2 with the help of stoichiometry.

To find grams of water(H2O) from grams of KNO2 , we need to follow three steps.

Step 1. Convert 2.47 grams of KNO2 to moles of KNO2.

$Moles = \frac{grams}{molar mass}$

Molar mass of KNO2 = 85.10 g/mol

$Moles = 2.47 gram KNO2\times \frac{1 mol KNO2}{85.10 gram KNO2}$

Moles = 0.0290 mol KNO2

Step 2. Convert moles of KNO2 to moles of H2O using mole ratio.

Mole ratio are the coefficient present in front of the compound in the balanced equation.

Mole ratio of KNO2 : H2O is 5 : 3 (5 coefficient of KNO2 and 3 coefficient of H2O)

$0.0290 mol KNO2\times \frac{3 mol H2O}{5 mol KNO2}$

Mole = 0.0174 mol H2O

Step 3. Convert mole of H2O to grams of H2O

Grams = Moles X molar mass

Molar mass of H2O = 18.00 g/mol

$Grams = 0.0174 mol H2O\times \frac{18 g H2O}{1 mol H2O}$

Grams of water = 0.313 grams H2O

Summary : The above three steps can also be done in a singe setup as shown below.

$2.47 gram KNO2\times \frac{(1 mol KNO2)}{(85.10 gram KNO2)}\times \frac{(3 mol H2O)}{(5 mol KNO2)}\times \frac{(18.00 gram H2O)}{(1mol H2O)}$

In the above setup similar units get cancelled out and we will get grams of H2O as 0.313 grams water (H2O)

8 0

4 years ago