Why was it so important that the scientist investigating the atom be young and unattached to existing scientific thinking

Name the cycloalkanes with molecular formula c6h12 that have a 3-membered ring and two substituents.

Tanzania [10]

Answer:

1-ethyl-2-methyl cyclopropane.

Explanation:

The structure of the molecule will be as shown in the attached image.

The molecular formula of the compound is C₆H₁₂.

It has 3 membered ring with 3 C atoms and two substituents one of them with one C atom (methyl) and the other with 2 C atoms (ethyl).

The ring consist of 3 C atoms, so its name is cyclo propane.

We numbering the atoms of the ring that give the ethyl substituent the low no. (1) and then methyl group take no. (2).

<em>Thus, the name of the compound is 1-ethyl-2-methyl cyclopropane.</em>

3 0

3 years ago

If an automobile air bag has a volume of 11.5 l , what mass of nan3 (in

True [87]

Moles of NaN3 at STP = volume of gas / 22.4 = 11.5/22.4 = 0.5mole. Massof NaN3 = moles of NaN3 x molecular weight = 0.5 x 65 = 32.5 grams.

8 0

3 years ago

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5 Questions to answer please

ryzh [129]

1.Type of bonding in which electrons are completely transferred is called ionic bond.
2. Isotopes have same atomic number but different atomic mass number.
Atomic number = number of protons + number of neutrons
Therefore, A is correct.
3. Nucleus is composed of neutrons and protons.
4. Chemical reactions follows the law of conservation of mass. Therefore mass of reactant = mass of product = 4 grams.
5. Again, mass of table salt formed should be equal to mass of (Na+Cl₂) = 4 grams.

5 0

3 years ago

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To what volume in millimeters must 50.0 mL of 18.0 M H2SO4 be diluted to obtain 4.35 M H2SO4?

poizon [28]

We know that to relate solutions of with the factors of molarity and volume, we can use the equation: $M_{1} V_{1} = M_{2} V_{2}$

**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.

So now we can assign values to these variables. Let us say that the 18 M $H_{2} SO_{4}$ is the left side of the equation. Then we have:

$(18 M)(0.050 L)=(4.35M) V_{2}$

We can then solve for $V_{2}$ :

$V_{2}= \frac{(18M)(0.05L)}{4.35M}$ and $V_{2} =0.21 L$ or $210 mL$

We now know that the total amount of volume of the 4.35 M solution will be 210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.

7 0

3 years ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

7 0

3 years ago