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2 years ago

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someone pls help me with my chemistry test plsss my teacher changes the questions so I can't search them up. its 21 questions so

I have to upload multiple posts but look at my account to get the rest pls.

1 answer:

tiny-mole [99]2 years ago

5 0

Answer:l

Explanation:

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Which are different forms of an element that have different numbers of neutrons?

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The number of neutrons is variable, resulting in isotopes, which are different forms of the same atom that vary only in the number of neutrons they possess. Together, the number of protons and the number of neutrons determine an element's mass number.

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2 years ago

Read 2 more answers

For the Bradford assay, the instructor will make a Bradford reagent dye by mixing 50 ml of 95% v/v ethanol with 100 mg of Coomas

Flauer [41]

Answer:

4.25% is the final concentration of phosphoric acid.

Explanation:

Initial concentration of phosphoric acid = $C_1=85\%=0.85$

Initial volume of phosphoric acid = $V_1=50 mL$

Final concentration of phosphoric acid = $C_2=?$

Final volume of phosphoric acid = $V_2=1 L=1000 mL$

( 1L = 1000 mL)

$C_1V_1=C_2V_2$

$C_2=\frac{C_1times V_1}{V_2}$

$=\frac{0.85\times 50 mL}{1000 mL}=0.0425=4.25%$

4.25% is the final concentration of phosphoric acid.

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2 years ago

Question 10 (4 points)

Alex17521 [72]

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The forces that hold atoms make compounds and molecules

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2 years ago

What processes can transform metamorphic rock to sediment.

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2 years ago

Suppose 7.05 g of zinc iodide is dissolved in 150. mL of a 0.20M aqueous solution of potassium carbonate. Calculate the final mo

sasho [114]

Answer:

$M=0.294M$

Explanation:

Hello!

In this case, since the molarity of the iodide ion is computed by dividing the moles of iodide ions by the volume of solution in liters (0.150 L), we first need to compute the moles of zinc iodide (ZnI2) by using its molar mass and then using a mole ratio of 1 mol of zinc iodide to 2 mol of iodide ion:

$n_{I^-}=7.05gZnI_2*\frac{1molZnI_2}{319.22 gZnI_2}*\frac{2molI^-}{1molZnI_2}=0.0442molI^-$

Therefore, the resulting molarity turns out:

$M=\frac{0.0442molI^-}{0.150L} \\\\M=0.294M$

Best regards!

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2 years ago