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3 years ago

If 3.64g of sand is separated from 5.25g of a mixture of salt and sand what is the percentage of sand in the sample

1 answer:

7 0

Total weight of the mixture of sand and salt = 5.25 g
Weight of sand separated = 3.64 g
Weight of the salt in the mixture = (5.25 - 3.64) g
= 1.61 g
Then the percentage of sand in the sample = [(3.64/5.25) * 100] percent
= 69.33%
So the percentage of sand in the total mixture of sand and salt is 69.33%.

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Which of the following represents the relative charge of proton

Snowcat [4.5K]

You’ve not put anything below so I’m not sure what the options are but the relative charge of a proton is +1

8 0

3 years ago

Write a balanced half-reaction for the reduction of bismuth oxide ion to bismuth ion in basic aqueous solution. Be sure to add p

Butoxors [25]

Answer:

$3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-$

Explanation:

Hello there!

In this case, according to the required half-reaction, we start by setting it up from bismuth (V) oxide ion to bismuth (III) ion:

$BiO_3^-\rightarrow Bi^{3+}$

Thus, next realize that the oxidation state of Bi in BiO3^- is 5+ because oxygen is 2- (-2*3+x=-1;x=-1+6;x=+5), so we obtain:

$(Bi^{5+}O_3)^-\rightarrow Bi^{3+}$

Thereafter, we realize three water molecules are needed on the right in order to balance the oxygens and consequently 6 hydrogen atoms on the left to balance hydrogen:

$6H^++(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O$

Now, since the balance is is basic media, we add six molecules of hydroxide ions in order to produce water with the hydrogen ones:

$6OH^-+6H^++(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O+6OH^-\\\\6H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O+6OH^-\\\\6H_2O-3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-$

Then, we accommodate the waters to obtain:

$3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-$

Best regards!

7 0

3 years ago

How to convert 120 barrels to quarts?

swat32

120 barrels = 15120 quarts
To convert from barrels to quarts, you just need to multiply the value in barrels by 126 (the conversion factor).
So, 120 barrels = 120 × 126 = 15120 quarts.

8 0

3 years ago

Read 2 more answers

What is the percent yield for a process in which 10.4g of CH3OH reacts and 10.1 g of CO2 is formed

monitta

Answer:

A. 70.7%

Explanation:

In the first step lets compute the molar mass of CH₃OH and CO

Molar Mass of CH₃OH = 1(12.01 g/mol) + 4(1.008 g/mol) +1(16.00 g/mol)

= 32.042 g/mol

Molar Mass of CO₂ = 1(12.01 g/mol) + 2(16.00 g/mol)

= 44.01 g/mol

Mass of only one reactant i.e. CH₃OH is given so it must be the limiting reactant. Next, the theoretical yield is calculated directly as follows:

Given mass of CH₃OH is 10.4 g. So we have:

10.4g CH₃OH

Convert grams of CH₃OH to moles of CH₃OH utilizing molar mass of CH₃OH as:

1 mol CH₃OH / 32.042 g CH₃OH

Convert CH₃OH to moles of CO₂ using mole ratio as:

2 mol CO₂ / 2 mol CH₃OH

Convert moles of CO₂ to grams of CO₂ utilizing molar mass of CO₂ as:

44.01 g/mol CO₂ / 1 mol CO₂

Now calculating theoretical yield using above steps:

[ 10.4 g CH₃OH ] [1 mol CH₃OH / 32.042 g CH₃OH ] [2 mol CO₂ / 2 mol CH₃OH] [44.01 g/mol CO₂ / 1 mol CO₂]

Multiplication is performed here. We are left with 10.4 and 44.01 g CO₂ from numerator terms in the above equation and 32.042 from denominator terms after cancellation process of above terms. So this equation becomes:

= ( 10.4 ) ( 44.01 ) g CO₂ / 32.042

= 457.704/32/042

= 14.28 g CO₂

Theoretical yield = 14.28 g CO₂

Finally compute the percent yield for a process in which 10.4g of CH₃OH reacts and 10.1 g of CO₂ is formed:

percent yield = (actual yield / theoretical yield) x 100

As we have calculated theoretical yield which is 14.28 g CO₂ and actual yield is 10.1 g CO₂ So,

percent yield = (10.1 g CO₂ / 14.28 g CO₂) x 100%

= 0.707 x 100%

= 70.7 %

Hence option A 70.7% yield is the correct answer.

8 0

4 years ago

Find the empirical formula of the following compounds:

Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus $\frac{mol phosphorus}{ 30.97 g phosphorus}$ = 0.0293 mol

Moles bromine 6.99 g bromine $\frac{mol bromine}{79.90 g bromine}$ =0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

brainly.com/question/14044066

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8 0

1 year ago