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Softa [21]
3 years ago
14

Which physical property does the inventor need to consider when choosing a

Chemistry
1 answer:
adelina 88 [10]3 years ago
6 0

Answer:

A.

............

Explanation:

Ductility

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Condensation 212F or 100C, Freezing 32F or 0C

Explanation:

Condensation 212 degrees Fahrenheit or 100 degrees Celsius.

Freezing point 32 degrees Fahrenheit or 0 degrees Celsius.

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Why is red the last color seen at sunset?
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Read 2 more answers
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
2 years ago
PLEASE HELP ME ASAPPPP
sattari [20]
Boyle’s law gives the relationship between pressure and volume of gases. It states that at constant temperature the pressure of gas is inversely proportional to volume of gas.
PV = k
Where P is pressure V is volume and k is constant
P1V1 = P2V2
Parameters at STP are on the left side and parameters for the second instance are on the right side of the equation
P1 - standard pressure - 1.0 atm
Substituting the values in the equation
1.0 atm x 5.00 L = P x 15.0 L
P = 0.33 atm
New pressure is 0.33 atm
5 0
3 years ago
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