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irina1246 [14]
2 years ago
14

Please help me out i'm new here but i really need help for science (if you need help with any hw like ela i can help!) (50 point

s) ima give more out in my other questions this is the 2 question make sure go to the first one!
Question 6 Slide 8 Pg 153 (image 1)
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Question 7 Slide 8 Pg 153 * (image 1)
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Question 8 Slide 8 Pg 153 * (image 1)
State the independent variable in this investigation?

Question 18 Slide 11 Pg 158 (image 2)

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Question 19 Slide 11 Pg 158 (image 2)
Is it...?? Choose one (check the image to see the questions and question answers)
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Question 20 Slide 11 Pg 158 (image 2)

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Chemistry
1 answer:
andrezito [222]2 years ago
6 0
Q6. 3
Q7. 3
Q8. pH
Q18. 3
Q19. 3
Q20. 4
Hope this helped??
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How many molecules are there in 10 grams of N2I6?
Anni [7]
10g/ 763g/mol= 0.013moles

6 0
2 years ago
Problem PageQuestion A chemist makes of magnesium fluoride working solution by adding distilled water to of a stock solution of
igomit [66]

Answer:

5.37 × 10⁻⁴ mol/L

Explanation:

<em>A chemist makes 660. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a 0.00154 mol/L stock solution of magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.</em>

Step 1: Given data

  • Initial concentration (C₁): 0.00154 mol/L
  • Initial volume (V₁): 230. mL
  • Final concentration (C₂): ?
  • Final volume (V₂): 660. mL

Step 2: Calculate the concentration of the final solution

We want to prepare a dilute solution from a concentrated one. We can calculate the concentration of the final solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁ / V₂

C₂ = 0.00154 mol/L × 230. mL / 660. mL = 5.37 × 10⁻⁴ mol/L

5 0
3 years ago
The Ksp can be used to find the concentration of hydroxide ions, and thus determine the pH given a concentration of of aluminum.
Anton [14]

Answer:

3.9.

Explanation:

(Molar mass of 342.15 g/mol).

The first thing to do in this question is to coverts our units. We are given 6.70 lb = 453.592 × 6.70 lb = 3039.0664 g of aluminum sulfate.

2050 gallons of water = 2050 × 3.785 = 7759.25 Litres.

After this we will now look for the Concentration aluminum sulfate using the formula below:

Concentration = number of moles ÷ volume. ------------------------------------(1).

Recall; Number of moles = 3039.0664/ 342.15 = 8.8823 moles. The volume = 2050 gallons = 7759.25 Litres.

Therefore, slotting in the values of of number of moles and volume into the equation (1) above, we have;

Concentration= 8.8823 moles/ 7759.25 Litres. = 0.001145 M.

The next equation we are going to make use of today is that of the solubilty product that is;

Ksp = (Al^3+) (OH^-)^3.

So, we have the ksp = 1.3 × 10^-33 and the value of (Al^3+)= 0.001145 M.

Hence, making (OH^-) the subject of the formula;

(OH^-) = ( 1.3 × 10^-33 / 0.001145)^1/3.

(OH^-) = 8 × 10^-11 M.

Hence, pOH = - log (OH^-)

pOH = - log (8 × 10^-11).

pOH = 10.1

Therefore, the values of pH is; 1

pH + pOH = 14.

pH = 14 - 10.1 = 3.9.

3 0
3 years ago
Zinc and magnesium metal each reacts with hydrochloric acid to make chloride salts of the respective metals, and hydrogen gas. a
kirill115 [55]
M=11.20 g
m(H₂)=0.6854 g
M(H₂)=2.016 g/mol
M(Mg)=24.305 g/mol
M(Zn)=65.39 g/mol
w-?

m(Mg)=wm
m(Zn)=(1-w)m

Zn + 2HCl = ZnCl₂ + H₂
m₁(H₂)=M(H₂)m(Zn)/M(Zn)=M(H₂)(1-w)m/M(Zn)

Mg + 2HCl = MgCl₂ + H₂
m₂(H₂)=M(H₂)m(Mg)/M(Mg)=M(H₂)wm/M(Mg)

m(H₂)=m₁(H₂)+m₂(H₂)
m(H₂)=M(H₂)(1-w)m/M(Zn)+M(H₂)wm/M(Mg)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}

m(H₂)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}

(1-w)/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}

1/M(Zn)-w/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}

w(1/M(Mg)-1/M(Zn))=m(H₂)/{M(H₂)m}-1/M(Zn)

w=[m(H₂)/{M(H₂)m}-1/M(Zn)]/(1/M(Mg)-1/M(Zn))

w=0.583 (58.3%)
5 0
3 years ago
If 8.00 mol of an ideal gas at stp were confined to a cube, what would be the length in cm of an edge of this cube?
MatroZZZ [7]
To solve this problem it is fundamentally, just look for the volume of the gas and convert it to cm3. At STP 1 mole = 22.4 liters. 8.00 moles x 22.4 liters/mole = 179.2 liters = 179,200 cm^3 Then. get the cube root of 179,200 cm^3. This would be equal to 56.38 cm and thus would be the length of the edge of this cube.
6 0
3 years ago
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