Please help 30 points and brainlyist!!!! All of the following are isotopes of hydrogen, except: A. 1/0 H
B. 1/1 H
C. 0/1 H
D. 2/1 H

Answers

Answer 1

Answer: My dear friend ✌ ✌

The isotopes of the hydrogen atom are-:

▶1 , 2 and 3

So,
According to your options,

➡Only C is wrong.

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Two very stable compounds, Freon-12 and Freon-14, are used as liquid refrigerants. A Freon-12 molecule consists of one carbon atom, two chlorine atoms, and two fluorine atoms. A Freon-14 molecule consists of one carbon atom and four fluorine atoms.81 In the space in your answer booklet, draw a structural formula for Freon-12. [1]

82 To which class of organic compounds do Freon-12 and Freon-14 belong? [1]

Answers

81. There is 1 carbon, 2 chlorine and fluorine atoms in Freon 12. To draw them it forms a cross with C in the middle and Cl and F both on the opposite side.
Cl
l
F - C- F
l
Cl

82. Freon-12 and Freon-14 are called halocarbons or just halides.

Answer:

- The structural formula for Freon-12 (CCl2F2)

Cl - C - Cl

It has the same structure as methane (CH4) with Hydrogen atoms replaced by Chlorine and Fluorine

- Freon-12 and Freon-14 belong halocarbons, a compound in which the Hydrogen of a hydrocarbon is replaced by halogens, like Chlorine and Fluorine.

How many grams of CuSO4 are there in 100.0 g of hydrate? How many moles?

Answers

the hydrate form of CuSO4 has 5 water molecules (CuSO4-5H20) copper (II) Sulfate pentahydrate or commonly known as blue vitriol.

To solve, the following molar masses are to be known.
CuSO4.5H2O (hydrate) - 249.7g/mole
CuSO4 (anhydrous) -159.6g/mole
Also there molar ratio of the hydrate and CuSO4 is 1.

the mass of the hydrate is to be divided by the molar mass of the hydrate then multiplied by the ratio (1) to get the moles of hydrate and multiplied by the molar mass of the anhydrous to get the mass in grams.

moles = (100g/249.7)*1 = 0.4 moles hydrate
grams = 0.4*159.6 = 64.9 grames hydrate

Andy is waiting at the signal. As soon as the light turns green, he accelerates his car at a uniform rate of 8.00 meters/second2. What is the displacement of Andy’s car after 30.0 seconds?A.
1200 meters
B.
2400 meters
C.
3600 meters
D.
4800 meters
E.
7200 meters

Answers

The answer is B
8m/s
Your going to multiply 8 and 30 together which gives you 240.

if you have 3.50 moles of hydrogen and 5.00moles of nitrogen to produce ammonia, witch element is the reactant in excess? 3H2+N2=2NH3

Answers

Given 3.50 moles of hydrogen and 5.00 moles of nitrogen to produce ammonia, nitrogen is the excess reactant.

What is the excess reactant?

The excess reactant is the reactant in a chemical reaction with a greater amount than necessary to react completely with the limiting reactant.

Step 1: Write the balanced equation.

3 H₂ + N₂ ⇒ 2 NH₃

Step 2: Establish the theoretical ratio.

The theoretical ratio (TR) of H₂ to N₂ is 3:1.

Step 3: Establish the experimental ratio.

The experimental ratio (ER) of H₂ to N₂ is 3.50:5.00 = 0.70:1.

Step 4: Determine the excess reactant.

Comparing TR and ER, we can realize that there is not enough hydrogen to react with the nitrogen. Thus, nitrogen is the excess reactant.

Given 3.50 moles of hydrogen and 5.00 moles of nitrogen to produce ammonia, nitrogen is the excess reactant.

Learn more about excess reactant here: brainly.com/question/17199947

Answer : The $N_2$ element is the reactant in excess.

Solution : Given,

Moles of $N_2$ = 5 moles

Moles of $H_2$ = 3.50 moles

The balanced chemical reaction is,

$3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

From the balanced reaction we conclude that

As, 3 moles of $H_2$ react with 1 mole of $N_2$

So, 3.5 moles of $H_2$ react with $(3.5)/(3)=1.16$ moles of $N_2$

The excess of $N_2$ = 5 - 1.16 = 3.84 moles

That means in the given balanced reaction, $H_2$ is a limiting reagent because it limits the formation of products and $N_2$ is an excess reagent.

Hence, the $N_2$ element is the reactant in excess.

The chemical equation below shows the photosynthesis reaction. 6CO2 + 6H2O mc010-1.jpg C6H12O6 + 6O2 The molar mass of carbon dioxide (CO2) is 44.01 g/mol. The molar mass of water (H2O) is 18.01 g/mol. A reaction uses 528 g of CO2. How many moles of water are used in this reaction?

Answers

Answer:

12 mol of water

Explanation:

$6CO_2 + 6H_2O \longrightarrow C_6H_(12)O_6 + 6O_2$

First we must find the moles of CO2.

We know that 1 mole of co2 has a mass of 4.01 g so how many moles will there be in 528 g.

We apply a simple rule of three

$44.01 g CO_2\longrightarrow 1 mol CO_2\n 528 g CO_2\longrightarrow x\nx= 528/44.01\nx=12 molCO_2$

By stoichiometry we know that for every 6 moles of carbon dioxide 6 moles of water are needed, now if we have 12 moles of carbon dioxide how many moles of water will be needed

We apply a simple rule of three

$6 molCO_2 \longrightarrow 6 mol H_2O\n 12 mol CO_2 \longrightarrow x\nx= ((12).(6))/(6)\n x= 12 mol H_2O$

Given:
6CO2 + 6H2O -> C6H12O6 + 6O2
molar mass of carbon dioxide (CO2) is 44.01 g/mol
molar mass of water (H2O) is 18.01 g/mol
528 g of CO2

Required:
moles of water

Solution:

528 g of CO2/ (44.01 g/mol CO2) = 12 moles CO2

12 moles CO2 (6 moles H2O/ 6 moles CO2) = 12 moles H2O

12 moles H2O (18.01 g/mol H2O) = 216.12 grams H2O

A 4.00 mL aliquot of a 0.15 M HCl solution is diluted to a final volume of 25.00 mL. What is the molarity of this first dilution solution? Then a second dilution was made by taking 7.50 mL of the first dilution and diluting it to 50.00 mL. What is the molarity of this second dilution?

Answers

Final answer:

The molarity of HCl after the first dilution is 0.024 M and after the second dilition, it is 0.0036M.

Explanation:

The problem at hand is a solution dilution problem in the field of Chemistry, usually tackled by the use of formula M1V1=M2V2, where M1 and V1 are the original molarity and volume, and M2 and V2 are the molarity and volume after dilution. In the first dilution, applying this formula gives (0.15 M)(4.00 mL) = (M2)(25.00 mL), solving for M2 gives a value of 0.024 M. The second dilution would similarly have the equation (0.024 M)(7.5 mL) = (M3)(50.0 mL), which gives M3 approximately 0.0036 M. Hence, the molarity of the first dilution would be 0.024 M and the molarity of the second dilution would be 0.0036 M.