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gogolik [260]
3 years ago
5

BRAINLIEST ASAP!!! Help

Chemistry
1 answer:
Andru [333]3 years ago
7 0

Answer:

N2C14

Explanation:

<em> determined the bond type by looking if it is a metal or nometal</em>

<em>Ionic Bond:NM+M</em>

<em>Covalent Bond:NM+NM</em>

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The class of organic compound below contains a carbonyl group as a part of its structure?
kolbaska11 [484]

Aldehydes and Ketones. Aldehydes and ketones are classes of organic compounds that contain a carbonyl (C=O) group.

that's the correct answer

can I get brainilest

5 0
2 years ago
How many moles are in 9.50 x 1022 molecules of CO? Be sure to ALL work below and report your answer using the correct number of
eimsori [14]

Answer:

0.158 moles

Explanation:

We are given;

9.50 x 10^22 molecules of CO

We are required to determine the number of moles;

We need to know;

1 mole of a compound = 6.022 × 10^23 molecules

Therefore;

9.50 x 10^22 molecules of CO will be equivalent to;

= 9.50 x 10^22 molecules ÷ 6.022 × 10^23 molecules/mole

= 0.158 moles

Therefore, the number of moles are 0.158 moles

4 0
3 years ago
Water is being pumped from the bottom of a well 150 feet deep at a rate of 200 gal/hour into a vented storage tank 30 feet above
KengaRu [80]

Explanation:

As the given data is as follows.

        Height, H = 150 feet

 Heat gain = 30,000 BTU/hr,  and  Heat loss = 25000 BTU/hr

  m = mass of water heated = 700 gallons = 5810 lbs

C_{p} is the heat capacity of water = 1 BTU/lb ^{o}F (given)

      \Delta T = temperature difference = 120^{o}F - 35^{o}F

Heat energy required to heat 700 gal can be calculated as follows:

    Heat Required = 5810 lbs \times 1 BTU/lb^{o}F \times (120^{o}F - 35^{o}F)

Thus, water rises till 120^{o}F.

3 0
3 years ago
The vapor pressure of pure water at 25 °c is 23.8 torr. What is the vapor pressure (torr) of water above a solution prepared by
maxonik [38]

The vapour pressure of the solution is 23.4 torr.

Use <em>Raoult’s Law</em> to calculate the vapour pressure:  

<em>p</em>₁ = χ₁<em>p</em>₁°  

where  

χ₁ = the mole fraction of the solvent  

<em>p</em>₁ and <em>p</em>₁° are the vapour pressures of the solution and of the pure solvent  

The formula for vapour pressure lowering Δ<em>p</em> is  

Δ<em>p</em> = <em>p</em>₁° - <em>p</em>₁  

Δ<em>p</em> = <em>p</em>₁° - χ₁<em>p</em>₁° = p₁°(1 – χ₁) = χ₂<em>p</em>₁°  

where χ₂ is the mole fraction of the solute.  

<em>Step 1</em>. Calculate the <em>mole fraction of glucose </em>

<em>n</em>₂ = 18.0 g glu × (1 moL glu/180.0 g glu) = 0.1000 mol glu  

<em>n</em>₁ = 95.0 g H_2O × (1 mol H_2O/18.02 g H_2O) = 5.272 mol H_2O  

χ₂ = <em>n</em>₂/(<em>n</em>₁ + n₂) = 0.1000/(0.1000 + 5.272) = 0.1000/5.372 = 0.018 62  

<em>Step 2</em>. Calculate the <em>vapour pressure lowering</em>  

Δ<em>p</em> = χ₂<em>p</em>₁° = 0.018 62 × 23.8 torr = 0.4430 torr  

<em>Step 3</em>. Calculate the <em>vapour pressure</em>  

<em>p₁</em> = <em>p</em>₁° - Δ<em>p</em> = 23.8 torr – 0.4430 torr = 23.4 torr

3 0
3 years ago
The arrows represent tge movement of which substances
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The arrows represent the movement of starting substances
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