Ask question

Ask question

All categories

2 years ago

6

A bond formed when atoms share electrons is a(n) __________.

1 answer:

Amiraneli [1.4K]2 years ago

8 0

Answer:

I think it is a covalent bon

You might be interested in

A student dissolved 3.50g of copper (II) nitrate in water and mixed it with a solution of sodium carbonate. The student recovere

Arisa [49]

Answer: If the theoretical yield is 2.35g then the percent yield is 80.4%

Explanation:

Given: Actual yield = 1.89 g

Theoretical yield = 2.35 g

Formula used to calculate the percentage yield is as follows.

$Percent yield = \frac{actual yield}{theoretical yield} \times 100$

Substitute the values into above formula as follows.

$Percent yield = \frac{actual yield}{theoretical yield} \times 100\\= \frac{1.89}{2.35} \times 100\\= 80.4 percent$

Thus, we can conclude that if the theoretical yield is 2.35g then the percent yield is 80.4%

3 0

3 years ago

Liquids will boil at a higher temperature when they are at a lower elevation true or false ?

DENIUS [597]

Question: Liquids will boil at a higher temperature when they are at a lower elevation true or false ?

Answer: True

6 0

3 years ago

Ammonia and oxygen react to form nitrogen monoxide and water, like this:

olchik [2.2K]

4NH3+5O2 <=>4NO + 6H2O

Using the definition of Kp, we have
Kp=(Pno^4*Ph2o^6)/(Pnh3^4*Po2^5)
where Pno=partial pressure of NO, etc.

The numerical value for a given temperature can be evaluated when the actual partial pressures are known.

6 0

3 years ago

Select all that are true statements for RNA*

Kitty [74]

Answer
RNA is single stranded

6 0

3 years ago

Read 2 more answers

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

<u>For process 2 :</u>

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago