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JulsSmile [24]
3 years ago
12

Can someone help me pls? Zr + O ⇒ B + O ⇒ Zn + C ⇒ Co + I ⇒ Mg + Br ⇒

Chemistry
1 answer:
Korolek [52]3 years ago
8 0

Explanation:

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Pure acetic acid (hc2h3o2) is a liquid and is known as glacial acetic acid. calculate the molarity of a solution prepared by dis
Alexandra [31]
In order to find the molarity of the solution, we first require the moles of acetic acid added. For this,we need the mass which is:

Mass = volume * density

Mass = 50 * 1.05
Mass = 52.5 grams


Moles = mass / molecular weight

Moles = 52.5 / 60.05 
Moles = 0.874 mol

Next, we know that the molarity of a solution is:

Molarity = moles / liter
Molarity = 0.874 / 0.5

Molarity = 1.75 M
3 0
3 years ago
Read 2 more answers
Which of the following water solutions would likely have the highest boiling point?
Vedmedyk [2.9K]

Answer:

1m

Explanation:

4 0
3 years ago
Over a long period of time, an igneous rock changes into a sedimentary rock. Which of these actions must have affected the igneo
Alla [95]

Answer: The actions that must have affected the igneous rock in order to form the sedimentary rock is that (It must have been broken down into sediments).

Explanation:

Rocks are solid structures that occurs naturally which is made up of different minerals. There are three main types of rocks, these includes:

--> METAMORPHIC ROCKS: These are the type of rocks which are formed by temperature and pressure changes inside the Earth.

--> SEDIMENTARY ROCKS: these rocks are usually formed from pre-existing rocks through the process of weathering (breaking down) of rocks.

--> IGNEOUS ROCKS: these rocks are formed when molten magma cools beneath or above the earth surface.

The actions that must have affected the igneous rock in order to form the sedimentary rock is that the igneous rocks are broken down into smaller pieces by erosion and weathering processes. Sediments which are formed accumulates at the earth surface. Over a long period of time, these sediments builds successive layers on top of one another. The sediments near the base hardens to form sedimentary rocks. This justifies the statement as a correct option (It must have been broken down into sediments).

3 0
3 years ago
Dilution question In many of the experiments, you will be asked to prepare a standard solution by diluting a stock solution. You
riadik2000 [5.3K]

Answer:

We could do two 1:50 dilutions and one 1:4 dilutions.

Explanation:

Hi there!

A solution that is 1000 ug/ ml  (or 1000 mg / l) is 1000 ppm.

Knowing that 1 ppm = 1000 ppb, 100 ppb is 0.1 ppm.

Then, we have to dilute the stock solution (1000 ppm / 0.1 ppm) 10000 times.

We could do two 1:50 dilutions and one 1:4 dilutions (50 · 50 · 4 = 10000). Since the first dilution is 1:50, you will use the smallest quantity of the stock solution (if we use the 10.00 ml flask):

First step (1:50 dilution):

Take 0.2 ml of the stock solution using the third dispenser (20 - 200 ul), and pour it in the 10.00 ml flask. Fill with water to the mark (concentration : 1000 ppm / 50 = 20 ppm).

Step 2 (1:50 dilution):

Take 0.2 ml of the solution made in step 1 and pour it in another 10.00 ml flask. Fill with water to the mark. Concentration 20 ppm/ 50 = 0.4 ppm)

Step 3 (1:4 dilution):

Take 2.5 ml of the solution made in step 3 (using the first dispenser 1 - 5 ml) and pour it in a 10.00 ml flask. Fill with water to the mark. Concentration 0.4 ppm / 4 = 0.1 ppm = 100 ppb.

6 0
3 years ago
Perform the calculation and record the answer with the correct number of significant figures.
kotykmax [81]

The question is incomplete, here is the complete question:

A. (6.5-6.10)/3.19

B. (34.123 + 9.60) / (98.7654 - 9.249)

<u>Answer:</u>

<u>For A:</u> The answer becomes 0.1

<u>For B:</u> The answer becomes 0.4884

<u>Explanation:</u>

Significant figures are defined as the figures present in a number that expresses the magnitude of a quantity to a specific degree of accuracy.

Rules for the identification of significant figures:

  • Digits from 1 to 9 are always significant and have infinite number of significant figures.
  • All non-zero numbers are always significant. For example: 664, 6.64 and 66.4 all have three significant figures.
  • All zeros between the integers are always significant. For example: 5018, 5.018 and 50.18 all have four significant figures.
  • All zeros preceding the first integers are never significant. For example: 0.00058 has two significant figures.
  • All zeros after the decimal point are always significant. For example: 2.500, 25.00 and 250.0 all have four significant figures.
  • All zeroes used solely for spacing the decimal point are not significant. For example: 10000 has one significant figure.

<u>Rule applied for addition and subtraction:</u>

The least precise number present after the decimal point determines the number of significant figures in the answer.

<u>Rule applied for multiplication and division:</u>

In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits

  • <u>For A:</u> (6.5-6.10)/3.19

This a a problem of subtraction and division.

First, the subtraction is carried out.

\Rightarow \frac{6.5-6.10}{3.19}=\frac{0.4}{3.19}

Here, the least precise number after decimal was 1.

\Rightarrow \frac{0.4}{3.19}=0.125

Here, the least precise number of significant digit is 1. So, the answer becomes 0.1

  • <u>For B:</u> (34.123 + 9.60) / (98.7654 - 9.249)

This a a problem of subtraction, addition and division.

First, the subtraction and addition is carried out.

\Rightarow \frac{34.123+9.60}{98.7654-9.249}=\frac{43.723}{89.5164}=\frac{43.72}{89.516}

Here, the least precise number after decimal in addition are 2 and in subtraction are 3

\Rightarrow \frac{43.72}{89.516}=0.48840

Here, the least precise number of significant digit are 4. So, the answer becomes 0.4884

6 0
3 years ago
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