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Juli2301 [7.4K]
3 years ago
11

What are the three possible combinations of the two types of

Chemistry
1 answer:
timurjin [86]3 years ago
6 0

Answer:

WHITE BLOOD SELL?

Explanation:

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According to the following equation:
Volgvan
2H2 + O2 → 2H2O
2 mol. → 2×(1×2+16)
1 mol. → x
x =  \frac{36 \times 1}{2 }  = 18 \: g \: \: of \:  h2o
6 0
3 years ago
If 251 mL of 6.0 M H2SO4 is diluted to 500. mL, what is its new molarity?
kolezko [41]

Answer:

  • <u>3.0M</u>

Explanation:

The dilution formula is:

       M_1\times V_1=M_2\times V_2

Where M₁ and M₂ are the molar concentrations and V₁ and V₂ are the volumes of the concentrated and the diluted solutions.

Substitute:

  • M₁ = <em>6.0M</em>
  • M₂ = unknown
  • V₁ = <em>251 mL</em>
  • V₂ = <em>500. mL</em>

         6.0M\times 251ml=M_2\times 500.ml

Clear M₂ and compute:

         M_2=6.0M\times 251ml/500.ml\approx3.0M

8 0
3 years ago
Two substances with empirical formula HNO are hyponitrous acid (M = 62.04 g/mol) and nitroxyl (M = 31.02 g/mol).(d) When hyponit
astra-53 [7]

The cis and Trans-forms of hyponitrous acid are given below in the attached document.

Hyponitrous acid has the chemical formula H2N2O2 or HON=NOH. 62.028 g/mol is the molecular weight of it. Additionally, it can take either a trans or cis form. When dry, the trans-hyponitrous acid crystallizes into white, explosive particles. Additionally, it has a half-life of 16 days at 25oC at a pH 1-3 in aqueous solution and is a weak acid (pKa1=7.21, pKa2=11.54). It decomposes into nitrous oxide and water. It is incorrect to think of N2O as the anhydride of H2N2O2. However, even though cis acid is unknown, we can still get its sodium salts.

Learn more about explosive particles here-

brainly.com/question/21437871

#SPJ4

5 0
2 years ago
What type of friction is talking
kotykmax [81]

Answer:

does friction talks ? friction is a oppossing force tho

4 0
3 years ago
Phosphoglucoisomerase interconverts glucose 6‑phosphate to, and from, glucose 1‑phosphate.
lawyer [7]

Answer:

Keq = 0.053

7.3 kJ/mol

Explanation:

Let's consider the following isomerization reaction.

glucose 6‑phosphate ⇄ glucose 1 - phosphate

The concentrations at equilibrium are:

[G6P] = 0.19 M

[G1P] = 0.01 M

The concentration equilibrium constant (Keq) is:

Keq = [G1P] / [G6P]

Keq = 0.01 / 0.19

Keq = 0.053

We can find the standard free energy change, ΔG°, of the reaction mixture using the following expression.

ΔG° = -R × T × lnKeq

ΔG° = -8.314 J/mol.K × 298 K × ln0.053

ΔG° = 7.3 × 10³ J/mol = 7.3 kJ/mol

7 0
4 years ago
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