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sladkih [1.3K]
3 years ago
7

Explain how the periodic table can help in writing electron configurations

Chemistry
1 answer:
Hoochie [10]3 years ago
4 0
Period : the no. of electron shells
group: the last no. of electron configuration
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How many moles of KNO3 are needed to make 600 ml of a 1.3M solution?
ludmilkaskok [199]
M = n / V

Where, M is molarity (M or mol/L), n is number of moles of the solute (mol) and V is volume of the solution (L).

Here the solute is KNO₃.
 The given molarity is 1.3 M
 This means 1L of solution has 1.3 moles of KNO₃.

Hence moles in 600 mL = 1.3 M x 0.6 L = 0.78 mol

Therefore to make 1.3 M KNO₃ solution, needed moles of KNO₃ is 0.78 mol
7 0
3 years ago
A man weighs 185 lb. What is his mass in grams?<br> Please show work.<br> Thank you
Vera_Pavlovna [14]

Answer:

83914.52 grams

Explanation:

Given that,

Weight of a man is 185 lb

We need to find his weight in grams

For this, we must know the relation between lb and grams.

We know that,

1 lb = 453.592 grams

To find the mass of man in grams, the step is :

185 lb = (453.592 × 185) grams

= 83914.52 grams

So, the mass of a man is 83914.52 grams.

8 0
4 years ago
650.0 mL of a gas is under a pressure of 840.0 torr. What would the volume of
zheka24 [161]

Answer:

Explanation:using the ideal gas formula, calculate the volume of 1.5omoles of a gas at 115kPA and a temperature of 298k?

6 0
2 years ago
Can someone help me with questions 8-12? What type of reaction is it?
statuscvo [17]
Isn't that a substitution reaction? Sorry if i am wrong haven't taken these in a while
5 0
3 years ago
PCL was mixed with gelatin to make a blend for elctrospun fibrous scaffold encapsulating growth factor that was admixed in the p
shusha [124]

Answer:

Explanation:

First PCL to form a scaffold, combined with gelatin.

They are made by first three forms A) made by just PCL.B) made by gelatin ratio PCL is 3:1 and last is C) made by gelatin PCL.

The decoration rate is 1%, 25% and 50% respectively.

<em>A) Growth factor is stuck in 21 days, and can not spread. In this case in vivo experiment for 7 days used highly degradable scaffold use the ability to break down due to decomposition in vivo degradation rate depends on the scaffold's acid byproduct impact.</em>

B) the amount of scaffolded degradation.

First, with scaffold A.

10 mg scaffold weight A= 1 per cent degradation.

Following degradation wt is 9.967=? Degradation per cent.

So, degradation (9.967* 1)/10= 0.9967 per cent.

Likewise for C) scaffold c 10 mg wt. Loss of 50 per cent.

After wt 8.33=? Degradation per cent.

Degradation (8.33* 50)/10=41.65 per cent.

Scaffold c degrades significantly, since the loss of wt is even greater.

7 0
4 years ago
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