All categories

3 years ago

What the different betwen observe and reference

1 answer:

7 0

Observe to to gather facts, by paying close attention towards what you are working on.

meanwhile, reference is the act or process on reaching to your conclusion, based on facts you already know .

You might be interested in

For the cells in a human body, an isotonic solution is 0.9% NaCl. If a red blood cell is placed in a 1% NaCl solution, what will

igomit [66]

Answer: The red blood cell will shrink due to water loss

Explanation:

Because 1% NaCl solution is slightly more concentrated than its isotonic 0.9% form, the red blood cell will lose its cell containing fluids such as water to the MORE concentrated environment.

This water loss after a prolonged period will result in shrinking of the red blood cell.

6 0

2 years ago

A student takes a sample of KOH solution and dilutes it with 100.00 mL of water. The student determines that the diluted solutio

Firlakuza [10]

Answer:

2.2mL

Explanation:

First, let us analyse what was given from the question:

C1 = 2.09M

V1 =?

C2 = 0.046M

V2 =100mL

Using dilution formula (C1V1 = C2V2), we can calculate the volume of the original solution as follows:

C1V1 = C2V2

2.09 x V1 = 0.046 x 100

Divide both side by the coefficient of V1 ie 2.09, we have:

V1 = (0.046 x 100) / 2.09

V1 = 2.2mL

8 0

3 years ago

Fritz Haber, a German chemist, discovered a way to synthesize ammonia gas (NH3) by combining hydrogen and nitrogen gases accordi

Westkost [7]

1) Write the balanced equation to state the molar ratios:

<span>3H2(g) + N2(g) → 2NH3(g)

=> molar ratios = 3 mol H2 : 1 mol N2 : 2 mol NH3

What volume of nitrogen is needed to produce 250.0 L of ammonia gas at STP?

First, convert the 250.0 L of NH3 to number of moles at STP .

Use the fact that 1 mole of gas at STP occupies 22.4 L

=> 250.0 L * 1mol/22.4 L = 11.16 L

Second, use the molar ratio to find the number of moles of N2 that produces 11.16 L of NH3

=> 11.16 L NH3 * [1 mol N2 / 2 mol NH3] = 5.58 mol N2

Third, convert 5.58 mol N2 into liters at STP

=> 5.58 mol N2 * [22.4 L/mol] = 124.99 liters

Answer: 124,99 liters

What volume of hydrogen is needed to produce 2.50 mol NH3 at STP?

First, find the number of moles of H2 that produce 2.50 mol by using the molar ratios:

2.50 mol NH3 * [3mol H2 / 2 mol NH3] = 3.75 mol H2

Second, convert the number of moles to liters of gas at STP:

3.75 mol * 22.4 L/mol = 84 liters of H2

Answer: 84 liters

</span>

6 0

3 years ago

Read 2 more answers

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti

koban [17]

Answer:

1) 2.0 g

2) 0 g

3) 4.17 g

4) 2.57 g

Explanation:

First of all, we need to know the compounds and the reaction. The ion carbonate is $CO3^{-2}$ , and the ion nitrate is $NO3^{-}$ .

Sodium is in group 1, so it must lose one electron to be stable, and be the cation $Na^{+}$ . Silver has only one electron too, so the cation will be $Ag^{+}$ .

To form the chemical compounds, first we put the cation, then the anion, and change their charges without the signal:

Sodium carbonate: Na2CO3

Silver nitrate: AgNO3

Silver carbonate: Ag2CO3

Sodium nitrate: NaNO3

The balanced reaction will be:

Na2CO3 + 2 AgNO3 --> Ag2CO3 + 2 NaNO3

Now, we must check the stoichiometry, which will be 1:2:1:2 (always in number of moles)

The question wants to know the mass value, so we need to know the molar mass of these compounds. Checking the periodic table will see that:

Na = 23 g/mol, C = 12 g/mol, N = 14 g/mol, O = 16 g/mol, Ag = 108 g/mol

So the molar mass of the compounds must be:

Na2CO3 = 106 g/mol (2x23 + 12 + 3x16)

AgNO3 = 170 g/mol (108 + 14 + 3x16)

Ag2CO3 = 276 g/mol (2x108 + 12 + 3x16)

NaNO3 = 85 g/mol

We have a mixture of the reactants, so one probably would be in excess, so, first will need to test. Let's do the stoichiometry calculus using silver nitrate as the limit, so:

1 mol of Na2CO3 ---------- 2 mol of AgNO3

106 g ------------------------------ 2x170 = 340 g

x ------------------------------------ 5.14 g

By a simple direct three rule:

340x = 544.84

x = 1.6 g of Na2CO3

That means that for this reaction, we only need 1.6 g of Na2CO3 to react with 5.14 of AgNO3. How we have 3.60 g of Na2CO3, it is on excess, and all the AgNO3 will be consumed.

1) The mass of Na2CO3 that remains after the reaction will be the initial less the mass that reacted:

m = 3.6 - 1. 6 = 2.0 g

2) All the AgNO3 reacted, so there isn't a mass present after the reaction.

m = 0 g

3) Now, doing the stoichiometry calculus between AgNO3 and Ag2CO3

2 moles of AgNO3 ------------- 1 mol of Ag2CO3

2x170 g ------------------------------- 276 g

5.14 g --------------------------------- x

By a simple direct three rule:

340x = 1418.64

x = 4.17 g of Ag2CO3

4) Now, doing the stoichiometry calculus between AgNO3 and NaNO3

2 moles of AgNO3 ----------------------- 2 moles of NaNO3

2x170 g ---------------------------------------- 2x85 g

5.14 g ------------------------------------------- x

By a simple direct three rule:

340x = 873.8

x = 2.57 g

8 0

2 years ago

What does he want? change into passive

Marina CMI [18]

There’s no picture if you meant to add a picture

3 0

3 years ago