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1 year ago

6

The figure represents a solid block of copper metal in a beaker at 30°C Which of the following best represents the particle arra

ngement as a student tips the beaker, which is kept at
30°C?

1 answer:

maria [59]1 year ago

6 0

Answer:

35

Explanation:

i think it is 30

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A 0.1153-gram sample of a pure hydrocarbon was burned in a C-H combustion train to produce 0.3986 gram of CO2and 0.0578 gram of

Mars2501 [29]

<u>Answer:</u> The mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2=0.3986g$

Mass of $H_2O=0.0578g$

Mass of sample = 0.1153 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.3986 g of carbon dioxide, $\frac{12}{44}\times 0.3986=0.1087g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.0578 g of water, $\frac{2}{18}\times 0.0578=0.0066g$ of hydrogen will be contained.

To calculate the percentage composition of a substance in sample, we use the equation:

$\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}\times 100$ ......(1)

<u>For Carbon:</u>

Mass of sample = 0.1153 g

Mass of carbon = 0.1087 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{0.1087g}{0.1153g}\times 100=94.27\%$

<u>For Hydrogen:</u>

Mass of sample = 0.1153 g

Mass of hydrogen = 0.0066 g

Putting values in equation 1, we get:

$\%\text{ composition of hydrogen}=\frac{0.0066g}{0.1153g}\times 100=5.72\%$

Hence, the mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

8 0

2 years ago

James was given a sample of sodium bicarbonate NaHCO3 in a weighing dish to analyze. His teacher asked him to find the percent o

gtnhenbr [62]

This question could be answered easily if the results of the abundance of the other elements are given. You will just have to subtract the sum of all their abundances to 100. Since it's not given, the solution would just be:

Na = 23 g/mol* 1 = 23 g
H = 1 g/mol * 1 = 1 g
C = 12 g/mol * 1 = 12 g
O = 16 g/mol * 3 = 48 g
Total = 84 g

% O = 48/84 * 100 = <em>57.14%</em>

5 0

2 years ago

Read 2 more answers

What is the volume of 43.7 g of helium at stp?

tankabanditka [31]

Answer is: volume of helium is 244.72 liters.
m(He) = 43.7 g.
n(He) = m(He) ÷ M(He).
n(He) = 43.7 g ÷ 4 g/mol.
n(He) = 10.925 mol.
V(He) = n(He) · n(He).
V(He) = 10.925 mol · 22.4 L/mol.
V(He) = 244.72 L.
Vm - molar volume at STP.
n - amount of substance.

0 0

2 years ago

Read 2 more answers

Identify the conjugate acid base pair <br> H3PO4(ag)+CO32=HCO3-(ag)+HPO42-(ag)

viktelen [127]

Answer:

H₃PO₄/H₂PO₄⁻ and HCO₃⁻/CO₃²⁻

Explanation:

An acid is a proton donor; a base is a proton acceptor.

Thus, H₃PO₄ is the acid, because it donates a proton to the carbonate ion.

CO₃²⁻ is the base, because it accepts a proton from the phosphoric acid.

The conjugate base is what's left after the acid has given up its proton.

The conjugate acid is what's formed when the base has accepted a proton.

H₃PO₄/H₂PO₄⁻ make one conjugate acid/base pair, and HCO₃⁻/CO₃²⁻ are the other conjugate acid/base pair.

H₃PO₄ + CO₃²⁻ ⇌ H₂PO₄⁻ + HCO₃⁻

acid base conj. conj.

base acid

3 0

2 years ago

What is the molarity of a HNO3 solution prepared by adding 290.7 mL of water to 350.0 mL of 12.3 M HNO3?

Lisa [10]

Answer:

6.72M of HNO3

Explanation:

In the problem you are diluting the original HNO3 solution by the addition of some water. The final volume is:

290.7mL + 350.0mL = 640.7mL

And you are diluting the solution:

640.7mL / 350.0mL = 1.8306 times

As the original concentration was 12.3M, the final concentration will be:

12.3M / 1.8306 =

<h3>6.72M of HNO3</h3>

5 0

2 years ago