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3 years ago

14

In the sentence "After conducting the science experiment,we recorded and analyzed the results," what does the word conducting me

an?

2 answers:

Dahasolnce [82]3 years ago

7 0

I agree performing probably

gulaghasi [49]3 years ago

4 0

Anything related to the word "performing/conducting"

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The solubility of nitrogen gas at 25 ◦C and 1 atm is 6.8×10−4 mol/L. If the partial pressure of nitrogen gas in air is 0.76 atm,

iris [78.8K]

Answer:

Concentration of dissolved nitrogen = 5.2 × 10⁻⁴ mol/L

Explanation:

More the pressure of the gas, more will be its solubility.

So, for two different pressure, the relation between them is shown below as:-

$\frac {P_1}{C_1}=\frac {P_2}{C_2}$

Given ,

P₁ = 1 atm

P₂ = 0.76 atm

C₁ = 6.8 × 10⁻⁴ mol/L

C₂ = ?

Using above equation as:

$\frac{1\ atm}{6.8\times 10^{-4}\ mol/L}=\frac{0.76\ atm}{C_2}$

$C_2=\frac{0.76\times 6.8\times 10^{-4}}{1}\ mol/L$

<u>Concentration of dissolved nitrogen = 5.2 × 10⁻⁴ mol/L</u>

5 0

2 years ago

Which of the following statements is true?

VikaD [51]

The answer is likely to be A

7 0

2 years ago

Read 2 more answers

26.0 g of a liquid that has a density of 1.44 g/mL needs to be measured out in a graduated cylinder. what volume should be used

Vaselesa [24]

As Density = Mass/Volume
Mass = 26.0g
Density = 1.44g/mL
Therefore Volume = Mass/Density
=> Volume = 26.0/1.44 = 18.055... = 18.1mL (to 3 sig figs)

4 0

2 years ago

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What is the electric field (in N/C) at a point 5.0 cm from the negative charge and along the line between the two charges?

Natali [406]

Answer: E = 2.455 x 10^5 N/C

Explanation:

q1 = 1.2x10^-7C

q2 = 6.2x10^-8C

Electric field, E = kQ/r²

where k = 9.0x10^9

since the location is (27 - 5)cm from q1

hence electric field, E1 = k*q1/r²

E1= (9x10^9 x 1.2x10^-7)/(0.22)² = 22314.05 N/C

for q2:

E1 = k*q2/r²

E2 at 5cm

E2 = (9x10^9 x 6.2x10^-8)/(0.05)² = 223200 N/C

Hence, the total electric field at 5cm position is

E = E1 + E2

E = 22314.05 + 223200 = 245514.05 N/C

E = 2.455 x 10^5 N/C

4 0

3 years ago

Suppose you wanted to dissolve 40.0 g NaOH in enough H2O to make 6.00 dm3 of solution

dezoksy [38]

Molarity of solution = 1.6 M

<h3>Further explanation</h3>

Given

40 g NaOH

6 L solution

Required

Steps to solve the problem of molarity

Solution

No additional information about the question.

If you want to make the solution above, then we just need to put the existing NaOH (40 g) into 6 L of water, then do the stirring (in a warm temperature above the hot plate will speed up the NaOH dissolving process)

But if you want to know the molarity of a solution, then

1. we calculate the moles of NaOH

$\tt mol=\dfrac{mass}{MW}$

MW(molecular weight) of NaOH=

Ar Na+ Ar O + Ar H

23 + 16 + 1 = 40 g/mol

so mol NaOH :

$\tt mol=\dfrac{40~g}{40~g/mol}=1~mol$

2. Molarity(M)

$\tt M=\dfrac{n}{V}\\\\M=\dfrac{1}{6}\\\\M=0.16$

5 0

2 years ago