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patriot [66]
2 years ago
11

Answer whats on the image PLEASE

Chemistry
1 answer:
kondaur [170]2 years ago
7 0

Answer:

1.97201398 g

Explanation:

Multiply the theoretical yield by the percent yield. Don't forget to move the decimal over 2 on the percent before multiplying or to use the percent button on your calculator.

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The ideal gas law equation is: PV = nRT
nignag [31]

Answer:

The answer is the first option 0.37 moles

Explanation:

the ideal gas law is PV = nRT, so in the terms of n,

n = PV/RT (divide both sides by RT to let n stand alone)

given: P = 700 kpa, V = 2L, T = 450 K , R = 8.31

n = (700 * 2) / (8.31 * 450)

= 1400/3739.5

= 0.34

8 0
2 years ago
In the formula for the compound XCl4, the X could represent<br> (1) C (3) Mg<br> (2) Η (4) Zn
Anvisha [2.4K]
Correct Answer: option 1 i.e. C

Reason: 
The the compound of interest i.e.  XCl4, since there are 4 Cl atoms bonded to X. This signifies that the valency of X is 4.

There atomic number of C is 6. It's electronic configuration is giving by 1s2 2s2 2p2. Thus, there are 4 electrons in valence shell of C. This signifies that valency of C is 4. Hence the compound present in present case is CCl4.
7 0
3 years ago
Read 2 more answers
The titration of Na2CO3 with HCl has the following qualitative profile: a. Identify the major species in solution at points A-F.
exis [7]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.

a) We need to find the major species from A to F.

Major Species at A:

1. Na_{2} CO_{3}

Major Species at B:

1. Na_{2} CO_{3}

2. NaHCO_{3}

Major Species at C:

1. NaHCO_{3}

Major Species at D:

1. NaHCO_{3}

2. H_{2}CO_{3}

Major Species at E:

1. H_{2}CO_{3}

Major Species at F:

1. H_{2}CO_{3}

b) pH calculation:

At Halfway point B:

pH = pKa_{1} + log[CO_{3}.^{-2}]/[HCO_{3}.^{-1}]

pH = pKa_{1} = 6.35

Similarly, at halfway point D.  

At point D,

pH = pKa_{2} + log [HCO_{3}.^{-1}]/[H2CO_{3}]

pH = pKa_{2} = 10.33

8 0
3 years ago
Your experiment requires 150 mL of 7.7 M NaOH. How many grams of NaOH will you need?
Elodia [21]
You have molarity and you have volume. Use the formula :
Molarity(M)= Moles(N)/Liter(L)            to get the solution. 
150 ml= .150 L
7.7 = N/.150
N=.1.155 moles of NaOH.
 And since you know the moles, use the molar mass to figure out the grams.
<span> (40g/mol NaOH) x (1.155mol) =  
46.2 g of NaOH.</span>
7 0
3 years ago
What kind of chemical bonds are represented by the product in the following diagram?
katovenus [111]
The chemical bonds in CH4 are all single bonds. C only can bond 4 times because it needs 8 electrons in it's outer shell and only has four right now. The bonds represented are all single bonds because there are two electrons present on each side of the carbon. Two electrons, in this case, equals one bond.
4 0
3 years ago
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