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allochka39001 [22]
2 years ago
7

Which among these were independent

Chemistry
1 answer:
prohojiy [21]2 years ago
3 0
Amount of sunlight, amount of water, and amount of fertilizer
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Need help on question 3 plz<br> URGENT HELP PLZ
Law Incorporation [45]
Alloys are the homogeneous mixture of metals and non metals.... they are used in making cars and jewellery to make them more featurable,,, means that they would possess high regidity to the destruction .... both features of metal and nonmetal
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3 years ago
Which type of reaction happens when a base is mixed with an acid? question 1 options:?
madreJ [45]
Chemicals cancel each other out and produce a salt and water. A neutralization reaction my also occur
7 0
3 years ago
Give the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by the element. rb
devlian [24]

Actually Rb or Rubidium in zero state has the following electron configuration:

<span>1s22s2</span><span>2p6</span><span>3s2</span><span>3p63d10</span><span>4s2</span><span>4p65s1</span>

 

However we can see that the ion has a 1 positive charge, which means that it lacks 1 electron, therefore the answer from the choices is:

<span>d. rb+: 1s22s22p63s23p64s23d104p6</span>
3 0
3 years ago
G= 10N/Kg
densk [106]

Answer: In classical physics terms, you do work on an object when you exert a force on ... One Newton is the force required to accelerate one kilogram of mass at 1 meter per second per second. ... The Newton-meters are termed joules (J). ... of the working object is transferred to that object raising its energy state.

Explanation:

7 0
3 years ago
You have 363 mL of a 1.25M potassium chloride solution, but you need to make a 0.50M potassium chloride solution. How many milli
maxonik [38]

Answer:- 544.5 mL of water need to be added.

Solution:- It is a dilution problem. The equation used for solving this type of problems is:

M_1V_1=M_2V_2

where, M_1 is initial molarity and  M_2 is the molarity after dilution. Similarly,  V_1 is the volume before dilution and  V_2 is the volume after dilution.

Let's plug in the values in the equation:

1.25M(363mL)=0.50M(V_2)

V_2=\frac{1.25M(363mL)}{0.50M}

V_2=907.5mL

Volume of water added = 907.5mL - 363mL  = 544.5 mL

So, 544.5 mL of water are need to be added to the original solution for dilution.

3 0
3 years ago
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