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tatuchka [14]
3 years ago
6

A bond that is formed when electrons are transferred from a metal atom to a nonmetal atom is a(n)

Chemistry
1 answer:
Daniel [21]3 years ago
3 0
Ionic bond hope it helped


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Complete and balance the following symbol equation, and add state symbols
erastovalidia [21]

Answer:

2NaOH + H2So4 》Na2So4+ 2H2O

Explanation:

SODIUM HYDROXIDE, 2NaOH IS (aq)

FULFURIC ACID, H2So4 IS (aq)

SODIUM SULFATE, Na2So4 IS (aq)

WATER,2H20 IS (l)

5 0
3 years ago
How is a shadow created during a lunar or solar eclipse?
Zepler [3.9K]
However, as a result of the Sun's large angular size, solar<span> illumination is only partially blocked in the outer portion of the Earth's </span>shadow<span>, which is given the name penumbra. A penumbral </span>eclipse<span> occurs when the Moon passes through the Earth's penumbra. The penumbra causes a subtle darkening of the Moon's surface.</span>
7 0
3 years ago
If the take off velocity of an airplane on a runway is 300 km /hr with an acceleration of 1 m/s2. What is the take off time of t
Rama09 [41]
<h3>Answer:</h3>

83.33 seconds.

<h3>Explanation:</h3>

<u>We are given;</u>

  • Take off velocity as 300 km/hr
  • Acceleration as 1 m/s²

We are required to calculate the take off time of the airplane.

<h3>Step 1: Convert velocity from km/hr to m/s </h3>

We are going to use the conversion factor.

The conversion factor is 3.6 km/hr per m/s

Therefore;

Velocity = 300 km/hr ÷ 3.6 km/hr per m/s

             = 83.33 m/s

<h3>Step 2: Calculate the take off time</h3>

We know that;

v = u + at

where, u is the initial velocity, v the final velocity, a the acceleration and t is time.

But, initial velocity is Zero

Therefore;

83.33 m/s = 1 m/s² × t

Thus;

time = 83.33 m/s ÷ 1 m/s²

       = 83.33 seconds

Therefore, the take off time is 83.33 seconds.

5 0
3 years ago
What is the definition if a chemical bond?
Anastaziya [24]
A. a mutual attraction between the nuclei and electrons in two different atoms.
8 0
3 years ago
Read 2 more answers
In yeast, ethanol is produced from glucose under anaerobic conditions. A cell‑free yeast extract is placed in a solution that co
nadezda [96]

Answer:

0.35 milli moles of ethanol can be theoretically be produced under these conditions.

Explanation:

C_6H_{12}O_6 + 2 ADP + 2 Pi\rightarrow 2 C_2H_5OH + 2 ATP + 2 CO_2

Moles of glucose = 3.50\times 10^2 milli mole

Moles of ADP = 0.35 milli mole

Moles of Pi = 0.35 milli mole

Moles of ATP = 0.70 milli mole

As we can see that ADP and Pi are in limiting amount which means tat they are limiting reagent. So, the moles of ethanol produced will depend upon the moles of ADP and Pi.

According to reaction, 2 moles of ADP gives 2 moles of glucose.

Then 0.35 milli moles of ADp will give :

\frac{2}{2}\times 0.35 mmol=0.35 mmol of ethanol

0.35 milli moles of ethanol can be theoretically be produced under these conditions.

5 0
3 years ago
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