Significant figures are an indicator of the certainty in measurements true or false?

Would the amount of solid magnesium increase decrease or stay the same if you added a strong acid to it?

Vlada [557]

Solid Magnesium is considered as active metal so it reacts with strong acids like HCl and H₂SO₄ liberating Hydrogen gas according to the following equations:

Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g)
Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g)

so the amount of solid magnesium decrease by addition of strong acid to it.

8 0

2 years ago

Use the picture below to answer the question. Comparing Wave A (blue) to Wave B (red), Wave B has a

fenix001 [56]

Answer: B longer wave

Explanation:

because

6 0

2 years ago

How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potass

sergiy2304 [10]

The amount of silver chromate that precipitates after addition of solutions is 12.44 g.

Number of moles:

The number of moles is the product of molarity of the solution and its volume. The formula is expressed as:

Moles = Molarity x Volume

Calculations:

Step 1:

The molecular formula of silver nitrate is AgNO3. The number of moles of silver nitrate is calculated as:

Moles of AgNO3 = 0.500 M x (150/1000) L

= 0.075 mol

Step 2:

The molecular formula potassium chromate is K2CrO4. The number of moles of potassium chromate is calculated as:

Moles of K2CrO4 = 0.400 M x (100/1000) L

= 0.04 mol

Step 3:

The balanced chemical reaction between AgNO3 and K2CrO4 is:

2AgNO3 + K2CrO4 -----> Ag2CrO4 + 2KNO3

The required number of moles of K2CrO4 = 0.075 mol/2 = 0.0375 mol

The given number of moles of K2CrO4 (0.04 mol) is more than the required number of moles (0.0375 mol). Therefore, AgNO3 is the limiting reagent.

Step 4:

According to the reaction, the molar ratio between AgNO3 and Ag2CrO4 is 2:1. Hence, the number of moles of Ag2CrO4 formed is 0.0375 mol.

The molar mass of Ag2CrO4 is 331.74 g/mol.

The mass of Ag2CrO4 is calculated as:

Mass = 0.0375 mol x 331.74 g/mol

= 12.44 g

Learn more about precipitation here:

brainly.com/question/13859041

#SPJ4

6 0

1 year ago

Find the natural abundance of Si-30.

nataly862011 [7]

<span> First you need to know how many isotopes there are of silicon, and its average atomic units (look at periodic table). Then make up a system of equations to solve for it. Theres 3 stable silicon isotopes (28, 29, 30) so you will need to have 3 equations. You must be given the percent abundance of at least one of the isotopes to solve because here I can only see 2 equations (numbered down below) set x = percent abundance of si-28 y = percent abundance of si-29 z = percent abundance of si-30 since all of silicon atoms account for 100% of all silicon: x + y + z = 100% = 1 therefore: 1) x = 1 - y - z You also have 2) 28x + 29y + 30z = average atomic mass you can substitute x so that equation becomes: 28 (1 - y - z) + 29y + 30z = average atomic mass See how you have 2 variables here? You cant go on until you know the value of one isotope already or you have given a clue which you can derive the third equation</span>

5 0

2 years ago

An antacid tablet with an active ingredient of CaCO3 was dissolved in 50.0 mL of 0.300 M HCl. When this solution was titrated wi

yaroslaw [1]

Answer:

Explanation:

mole of HCl remaining after reaction with CaCO₃

= .3 M of NaOH of 32.47 mL

= .3 x .03247 moles

= .009741 moles

Initial HCl taken = .3 x .005 moles = .0015 moles

Moles of HCl reacted with CaCO₃

= .009741 - .0015 = .008241 moles

CaCO₃ + 2HCl = CaCl₂ + CO₂ + H₂O .

1 mole 2 moles

2 moles of HCl reacts with 1 mole of CaCO₃

.008241 moles of HCl reacts with .5 x .008241 moles of CaCO₃

CaCO₃ reacted with HCl = .5 x .008241 = .00412 moles

the mass (in grams) of calcium carbonate in the tablet

= .00412 x 100 = .412 grams . ( molar mass of calcium carbonate = 100 )

5 0

2 years ago