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Advocard [28]
3 years ago
15

In a(n) _____ contract, the powerful party dictates the terms of the agreement and eliminates the other party’s free will.

Chemistry
1 answer:
Lesechka [4]3 years ago
6 0
I honestly think it would be a victor contract.. 


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What is an ionic bond in an electron?
Igoryamba
Its the complete transfer of valence electrons between atoms 
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3 years ago
A 1.53-L sample of gaseous sulfur dioxide has a pressure of 5600
andriy [413]
5600 why because you take away and then you add then yeah
5 0
3 years ago
Help me please!!
WINSTONCH [101]

Answer:

663 g

Explanation:

Step 1: Write the balanced equation

2 LiOH + CO₂ ⇒ H₂O + Li₂CO₃

Step 2: Calculate the moles corresponding to 825 L of CO₂

At standard pressure and temperature, 1 mole of CO₂ has a volume of 22.4 L.

825 L × 1 mol/22.4 L = 36.8 mol

Step 3: Calculate the moles of H₂O formed from 36.8 moles of CO₂

The molar ratio of CO₂ to H₂O is 1:1. The moles of H₂O formed are 1/1 × 36.8 mol = 36.8 mol.

Step 4: Calculate the mass corresponding to 36.8 moles of H₂O

The molar mass of H₂O is 18.02 g/mol.

36.8 mol × 18.02 g/mol = 663 g

3 0
2 years ago
For an object to remain at rest, which of the following must be true?
zubka84 [21]

Answer:

the forces on it are balanced

Explanation:

7 0
3 years ago
Read 2 more answers
To standardize an H2SO4 solution, you put 20.00 mL of it in a flask with a few drops of indicator and put 0.450 M NaOH in a bure
hram777 [196]

To standardize 20.00 mL of 0.495 M H₂SO₄ are used 44.10 mL of 0.450 M NaOH in a neutralization reaction.

We want to standardize a H₂SO₄ solution with NaOH. The neutralization reaction is:

H₂SO₄ + 2 NaOH ⇒ Na₂SO₄ + H₂O

The buret, which contains NaOH, reads initially 0.63 mL, and 44.73 mL at the endpoint. The used volume of NaOH is:

V = 44.73 mL - 0.63 mL = 44.10 mL

44.10 mL of 0.450 M NaOH are used for the titration. The reacting moles of NaOH are:

0.04410 L \times \frac{0.450mol}{L} = 0.0198 mol

The molar ratio of H₂SO₄ to NaOH is 1:2. The moles of H₂SO₄ that react with 0.0198 moles of NaOH are:

0.0198 mol NaOH \times \frac{1molH_2SO_4}{2molNaOH} = 0.00990 molH_2SO_4

0.00990 moles of H₂SO₄ are in 20.00 mL of solution. The molarity of H₂SO₄ is:

[H_2SO_4] = \frac{0.00990mol}{0.02000} = 0.495 M

To standardize 20.00 mL of 0.495 M H₂SO₄ are used 44.10 mL of 0.450 M NaOH in a neutralization reaction.

Learn more: brainly.com/question/2728613

7 0
3 years ago
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