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3 years ago

Gold has a density of about 20.0g/mL this means that each milliliter has a mass of about __ grams.

Chemistry

2 answers:

qwelly [4]3 years ago

8 0

Answer:

20.0 grams

Explanation:

If the density of gold is 20.0 g/mL, then we can multiply it by 1 mLto find the weight of 1 mL of gold.

20.0 $\frac{grams}{mL}$ *1mL=20.0 grams

dimaraw [331]3 years ago

4 0

Answer:

20.0 grams

Explanation:

If the density of gold is 20.0 g/mL, then we can multiply it by 1 mLto find the weight of 1 mL of gold.

20.0\frac{grams}{mL}*1mL=20.0 grams

Explanation:

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An actacide tablet containing Mg(OH)2 (MM = 58.3g / (mol)) is titrated with a 0.100 M solution of HNO3. The end point is determi

PtichkaEL [24]

Answer:

0.0583g

Explanation:

The equation of the reaction is;

2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)

From the question, number of moles of HNO3 reacted= concentration × volume

Concentration of HNO3= 0.100 M

Volume of HNO3 = 20.00mL

Number of moles of HNO3= 0.100 × 20/1000

Number of moles of HNO3 = 2×10^-3 moles

From the reaction equation;

2 moles of HNO3 reacts with 1 mole of Mg(OH)2

2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2

But

n= m/M

Where;

n= number of moles of Mg(OH)2

m= mass of Mg(OH)2

M= molar mass of Mg(OH)2

m= n×M

m= 1×10^-3 moles × 58.3 gmol-1

m = 0.0583g

6 0

3 years ago

Common additives to drinking water include elemental chlorine, chloride ions, and phosphate ions. Recently, reports of elevated

NeX [460]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

8 0

3 years ago

How does air bubbles affect density of a solid object?

kirill [66]

it decreases the density of the object the air bubbles take up space. it increases the volume of the object slightly but the objects weight remains the same, hence the objects density decreases

6 0

3 years ago

Calculate the following quantity: molarity of a solution prepared by diluting 45.45 mL of 0.0404 M ammonium sulfate to 550.00 mL

dybincka [34]

Answer:

$M_2=3.34x10^{-3}M$

Explanation:

Hello!

In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

$M_1V_1=M_2V_2$

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

$M_2=\frac{M_2V_2}{V_1} =\frac{45.45mL*0.0404M}{550.00mL}\\\\M_2=3.34x10^{-3}M$

Best regards!

5 0

2 years ago

A meteorologist filled a weather balloon with 3.00L of the inert noble gas helium. The balloon's pressure was 765 torr. The ball

yanalaym [24]

Answer:

4.33 L

Explanation:

Step 1: Given data

Initial volume of the balloon (V₁): 3.00 L

Initial pressure of the balloon (P₁): 765 torr

Final volume of the balloon (V₂): ?

Final pressure of the balloon (P₂): 530 torr

Step 2: Calculate the final volume of the balloon

If we consider Helium to behave as an ideal gas, we can calculate the final volume of the balloon using Boyle's law.

$P_1 \times V_1 = P_2 \times V_2\\V_2 = \frac{P_1 \times V_1}{P_2} = \frac{765torr \times 3.00L}{530torr} = 4.33 L$

7 0

3 years ago