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3 years ago

What is formed when 2 non metals reacts?

Chemistry

1 answer:

ryzh [129]3 years ago

4 0

Answer:

A) covalent bond

Explanation:

Covalent bonding generally happens between nonmetals.

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Which represents the converted length of the 205 meter long cruise ship into kilometers? A) 0.0205 km B) 0.205 km C) 2.05 km D)

zmey [24]

I believe the answer is A) 0.0205 km

7 0

3 years ago

Read 2 more answers

As a part of a clinical study, a pharmacist is asked to prepare a modification of a standard 22g package of a 2% mupirocin ointm

Vesna [10]

Answer:

226.8 mg of mupirocin powder are required

Explanation:

Given that;

weight of standard pack = 22 g

mupirocin by weight = 2%

so weight of mupirocin = 2% × 22 = 2/100 × 22 = 0.44 g

so by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment

mg of mupirocin powder are required = ?, lets rep this with x

Total weight of ointment = 22 + x g

Amount of mupirocin = 0.44 + x g

percentage of mupirocin in ointment is 3?

3/100 = 0.44 + x g / 22 + x g

3( 22 + x g ) = 100( 0.44 + x g )

66 + 3x g = 44 + 100x g

66 - 44 = 100x g - 3x g

97 x g = 22

x g = 22 / 97

x g = 0.2268 g

we know that; 1 gram = 1000 Milligram

so 0.2268 g = x mg

x mg = 0.2268 × 1000

x mg = 226.8 mg

Therefore, 226.8 mg of mupirocin powder are required

8 0

3 years ago

A missile is flying at a speed of 125 m/s. If the missile has a mass of 125 kg, what’s its kinetic energy?

LenKa [72]

976,563 is the answer I got to your question

5 0

2 years ago

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Fructose-1-P is hydrolyzed according to: Fructose-1-P + H2O → Fructose + Pi If a 0.2 M aqueous solution of Froctose-1-P is allow

aalyn [17]

Answer:

$\Delta G^{\circ}=-15902 J/mol$

Explanation:

In this problem we only have information of the equilibrium, so we need to find a expression of the free energy in function of the constant of equilireium (Keq):

$\Delta G^{\circ}=-R*T*ln(K_{eq})$

Being Keq:

$K_{eq}=\frac{[fructose][Pi]}{[Fructose-1-P]}$

Initial conditions:

$[Fructose-1-P]=0.2M$

$[Fructose]=0M$

$[Pi]=0M$

Equilibrium conditions:

$[Fructose-1-P]=6.52*10^{-5}M$

$[Fructose]=0.2M-6.52*10^{-5}M$

$[Pi]=0.2M-6.52*10^{-5}M$

$K_{eq}=\frac{(0.2M-6.52*10^{-5}M)*(0.2M-6.52*10^{-5}M)}{6.52*10^{-5}M}$

$K_{eq}=613.1$

Free-energy for T=298K (standard):

$\Delta G^{\circ}=-8.314\frac{J}{mol*K}*298K*ln(613.1)$

$\Delta G^{\circ}=-15902 J/mol$

3 0

3 years ago

PLEASE HELP ASAP A gardener plants a new kind of plant that will attract more helpful insects to his garden. Suddenly, the garde

stiks02 [169]

Answer:

Hope it helps you.....

The answer is in form of image..

See it...

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3 years ago