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alisha [4.7K]
2 years ago
6

What percentage of radioactive substance remains after two half-lives

Chemistry
1 answer:
Ksju [112]2 years ago
3 0

Answer:

After 2 half-lives there will be 25% of the original isotope, and 75% of the decay product. After 3 half-lives there will be 12.5% of the original isotope, and 87.5% of the decay product. After 4 half-lives there will be 6.25% of the original isotope, and 93.75% of the decay product.

Explanation:

You might be interested in
I need help with this!
bekas [8.4K]

Answer:

FILTRATION.

filter the solution and separate sand and salt solution

evaporation

evaporate the water and salt mixture and salt will be left as a solid

4 0
2 years ago
A beaker contains a 25 mL solution of an unknown monoprotic acid that reacts in a 1:1 stochiometric ratio with NaOH. Titrate the
Luden [163]

Answer:

concentration of HCL = 4.56 g/dm3

Explanation:

If the titrant and analyte have a 1:1 mole ratio, the formula is molarity (M) of the acid x volume (V) of the acid = molarity (M) of the base x volume (V) of the base. (Molarity is the concentration of a solution expressed as the number of moles of solute per litre of solution.)

Step 1: Calculate the amount of sodium hydroxide in moles

Volume of sodium hydroxide solution = 25.0 ÷ 1,000 = 0.0250 dm3

Rearrange:

Concentration in mol/dm3 = \frac{\textup{amount~of~solute~in~mol}}{\textup{volume~in~dm}^3}

Amount of solutein mol = concentration in mol/dm3 × volume in dm3

Amount of sodium hydroxide = 0.100 × 0.0250

= 0.00250 mol

Step 2: Find the amount of hydrochloric acid in moles

The balanced equation is: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

So the mole ratio NaOH:HCl is 1:1

Therefore 0.00250 mol of NaOH reacts with 0.00250 mol of HCl

Step 3: Calculate the concentration of hydrochloric acid in mol/dm3

Volume of hydrochloric acid = 20.00 ÷ 1000 = 0.0200 dm3

Concentration in mol/dm3 = \frac{\textup{amount~of~solute~in~mol}}{\textup{volume~in~dm}^3}

Concentration in mol/dm3 = \frac{\textup{0.00250}}{\textup{0.0200}}

= 0.125 mol/dm3

Step 4: Calculate the concentration of hydrochloric acid in g/dm3

Relative formula mass of HCl = 1 + 35.5 = 36.5

Mass = relative formula mass × amount

Mass of HCl = 36.5 × 0.125

= 4.56 g

So concentration = 4.56 g/dm3

8 0
2 years ago
How many grams sodium bromide can be formed from 51 grams of sodium hydroxide?
raketka [301]

Explanation:

When working with moles only, you will start by applying stoichiometry to determine how the reactants will affect your amount of products in this reaction. For this question, we will assume that other reactants are in infinite qualities, so therefore, it is the amount of aluminum that we will be concerned with. You need to figure out how much aluminum is in the specified amount of aluminum bromide, and then how much aluminum hydroxide that will be able to create. Make sure all your units cancel out!

9.24 mol AlBr3 x (1 mol Al / 1 mol AlBr3) x (1 mol Al(OH)3 / 1 mol Al) = 9.24 mol AlBr3

When you're working with mole ratios that involve grams to moles conversions, the first thing you want to do is calculate the molecular weight of each component you are being asked about. Because the question was given to you as words instead of chemical formulas, you will want to figure out the chemical formulas. For example, aluminum hydroxide is Al(OH)3 and aluminum bromide is AlBr3. To calculate molecular weight, you will want to consult a periodic table, find the molecular weight for each atom, and then calculate the correct sum of each molecular weight. Make sure you keep track of the number of each atom you have, i.e. 3 oxygen and 3 hydrogen for aluminum hydroxide.

Na = 22.990 g/mol

O = 15.999 g/mol

H = 1.008 g/mol

NaOH = 22.990 g/mol + 15.999 g/mol + 1.008 g/mol = 39.997 g/mol

Al = 26.982 g/mol

O = 15.999 g/mol

H = 1.008 g/mol

Al(OH)3 = 26.982 g/mol + (3 x 15.999 g/mol) + (3 x 1.008 g/mol) = 78.003 g/mol

Now, if you begin with an amount of NaOH in grams, you will first have to convert that to moles in order to use the mole ratio.

24 g NaOH x (1 mol NaOH / 39.997 g NaOH) = 0.600 mol NaOH

Now, you will have to account for the part of the sodium hydroxide that will be present in the aluminum hydroxide. In this case, it is the hydroxide (OH) portion of the formula. There is one mole of OH in each mole of NaOH, but there are 3 moles of OH in each mole of Al(OH)3. You will start with the 0.600 mol NaOH you know you have and then use the mole ratio.

0.600 mol NaOH x (1 mol OH / 1 mol NaOH) x (1 mol Al(OH)3 / 3 mol OH) = 0.200 mol Al(OH)3

Finally, when you are converting from grams to grams, you will have to find the molecular weight of both the reactant and the product, convert reactants in grams to reactants in moles, then use the mole ratio, then convert the moles of product back to grams of product. This time, you are concerned about the mole ratio of sodium, as that is the element that is in both chemical formulas.

7 0
2 years ago
If the reaction of 150. G of ammonia with 150. G of oxygen gas yields 87. G of nitric oxide (no), what is the percent yield of t
Gekata [30.6K]

The percentage yield of the given reaction is 77.33%.

What is percentage yield?
Reactants
frequently produce fewer product quantities than predicted by the chemical reaction's formula. The percentage of a theoretical yield that's been produced in a reaction is calculated using the percent yield formula. Working through a stoichiometry problem yields the theoretical yield, which is the ideal quantity of the final product. The actual yield is determined by calculating the volume of the product formed. We can determine the percentage yield by dividing the actual yield by the theoretical yield.

Moles is calculated by using the formula:
Moles of Ammonia:
Given mass of ammonia = 150g
Molar mass of ammonia = 17 g/mol
Putting values in above equation, we get:

Moles of Oxygen
Given mass of oxygen = 150g
Molar mass of oxygen = 32 g/mol
Putting values in above equation, we get:
For the given chemical equation:

By Stoichiometry,
5 moles of oxygen reacts with 4 moles of ammonia.
So, 4.6875 moles of oxygen will react with =  of ammonia
As, moles of ammonia required is less than the calculated moles. Hence, ammonia is present in excess and is termed as excess reagent.
Therefore, oxygen is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the given reaction:
5 moles of oxygen gas produces 4 moles of nitric oxide
So, 4.6875 moles of oxygen gas will produce =  of nitric oxide

Now, to calculate the theoretical amount of nitric oxide, we use equation 1: Molar mass of nitric oxide = 30 g/mol
 Given mass of nitric oxide = 112.5 g
Now, to calculate the percentage yield, we use the formula:
Experimental yield = 87 g
Theoretical yield = 112.5 g
Putting values in above equation, we get:
Hence, the percentage yield of the given reaction is 77.33%.

learn more about percentage yield
brainly.com/question/14714924
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6 0
1 year ago
Is the first part "kilo" from the word kilograms a suffix, prefix or unit???​
Nikolay [14]

Answer:

Kilo is a prefix

Explanation:

Prefixes are the beginning of words.

Thus KILOgram is at the beginning.

Suffix is at the END of a word

Unit is the body of the word

7 0
3 years ago
Read 2 more answers
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