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ohaa [14]
3 years ago
15

Convert 19,864 mm to m​

Chemistry
2 answers:
navik [9.2K]3 years ago
3 0
19.864 meters would be the correct answer
mart [117]3 years ago
3 0

I think 19,864 mm converted to m is 19.864 I'm not 100% sure tho

Hope this helps =w=

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What is the independent variable?<br> (For science)
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Answer:

The independent variable is what you change in the experiment.

Explanation:

The independent determines the dependent variable.

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What structure holds the sister chromatids to the spindle fibers?
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The answer is <span>a. kinetochore.

A kinetochore is a protein structure that holds the </span><span>sister chromatids to the spindle fibers. It is the place on chromatids where the spindle fibers bind during the cell division. As the result, sister chromatids are pulled apart to the opposite ends of the cell.</span>
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3 years ago
How many grams of potassium chloride are required to make 250 ml of a 0.75 m kcl solution?
DedPeter [7]

Answer:

The solution would need 13.9 g of KCl

Explanation:

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0.75 moles in 1 kg of solvent.

Let's think as an aqueous solution.

250 mL = 250 g, cause water density (1g/mL)

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250 g will have (0.75 . 250)/1000 = 0.1875 moles of KCl

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7 0
3 years ago
Jolene wants to experiment with sugar cubes. Which of the following causes a sugar cube to only change physically, not chemicall
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5 0
3 years ago
How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?
meriva

Answer:

\large \boxed{\text{77.4 mL}}

Explanation:

                Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

    V/mL:     249

c/mol·L⁻¹:  0.0443     0.285

1. Calculate the moles of Ba(OH)₂

\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1  mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}

3. Calculate the volume of HCl

V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}

8 0
3 years ago
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