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serious [3.7K]
2 years ago
6

Consider butter (density= 0.860 g/mL) and sand (density= 2.28 g/mL). If 1.00 mL of butter were mixed with 1.00 mL of sand and mi

xed as thoroughly as possible what would be the density of the resulting mixture?
Chemistry
1 answer:
KiRa [710]2 years ago
6 0

The density of the mixture will be 1.57 g/cm³.


Step 1. Calculate the <em>mass of the butter</em>.


\text{Mass} = \text{1.00 cm}^{3 } \times \frac{\text{0.680 g} }{\text{1 cm}^{3 }} = \text{0.860 g}\\

Step 2. Calculate the <em>mass of the sand</em>.


\text{Mass} = \text{1.00 cm}^{3 } \times \frac{\text{2.28 g} }{\text{1 cm}^{3 }} = \text{2.28 g}\\

Step 3. Calculate the <em>density of the mixture</em>.

Total mass = 0.860 g + 2.28 g = 3.14 g.

Total volume = 1 cm³ + 1 cm³ = 2 cm³

\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{\text{3.14 g} }{\text{2 cm}^{3 }} = \textbf{1.57 g/cm}{^{3}\\

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What is the change in enthalpy in kilojoules when 3.24 g of CH3OH is completely reacted according to the following reaction 2 CH
vodka [1.7K]

Answer:

12.78 kJ

Explanation:

The correct balanced reaction would be

2CH_3OH\rightarrow 2CH_4+O_2\Delta H=252.8\ \text{kJ}

Mass of methanol = 3.24\ \text{g}

Moles of methanol can be obtained by dividing the mass of methanol with its molar mass (32.04\ \text{g/mol})

\dfrac{3.24}{32.04}=0.10112\ \text{moles}

Enthalpy change for the number of moles is given by

\dfrac{\text{Number of moles of methanol in the reaction}}{\text{Enthalpy change in the reaction}}=\dfrac{\text{Number of moles in 3.24 g of methanol}}{\text{Enthaply in change in the mass of methanol}}

\\\Rightarrow\dfrac{2}{252.8}=\dfrac{0.10112}{\Delta H}\\\Rightarrow \Delta H=\dfrac{0.10112\times 252.8}{2}\\\Rightarrow \Delta H=12.781568\approx 12.78\ \text{kJ}

The change in enthalpy is 12.78 kJ.

5 0
2 years ago
It takes energy to ionize any atom. It takes progressively more and more energy for each successive electron that is removed and
IrinaVladis [17]

Answer:

Removal of Third Electron

Explanation:

a major jump is required to remove the third electron. In general, successive ionization energies always increase because each subsequent electron is being pulled away from an increasingly more positive ion.

Ionization energy increases from bottom to top within a group, and increases from left to right within a period.

5 0
2 years ago
How many moles of PBr3 contain 3.68 x 10^25 bromine atoms?
scoray [572]
<span>3.68 x 10²⁵ bromine atoms * 1mol/6.02*10²³ atoms=
 = 61.13 mol of bromine atoms

1 mol PBr3 ----- 3 mol Br
x mol PBr3 -----61.13 mol Br

x= 1*61.13/3 = 20.4 mol PBr3.


</span>20.4 mol PBr3 <span>contain 3.68 x 10^25 bromine atoms.</span>
7 0
2 years ago
Read 2 more answers
A piece of aluminum foil 1.00cm square and 0.590mm thick is allowed to react with bromine to form aluminum bromide. Part A How m
nalin [4]

Answer:

Part A: 5.899x10^-3 moles of Al

Part B: 1.573 g of AlBr3

Explanation:

Part A: We have to obtain the volume of the piece of aluminium; all sides of the square must be in cm. Then, use the density to obtain the mass.

0.590 mm  (\frac{1 cm}{10 mm}) = 0.059 cm

V= lxlxl= (1cm)(1cm)(0.059 cm)= 0.059 cm^3.

0.059 is the volume of the Al udes for the reaction. Now, to oabtain the moles:

0.059 cm^3 Al(\frac{2.699 g Al}{1 cm^3 Al})(\frac{1 mol Al}{27 g Al})= 5.899x10^-3 moles

Part B: To obatin the mass of AlBr3, we need the balanced chemical equation:

                                               2Al + 3Br2 → 2AlBr3

We assume bromine (Br2) is in excess, therefore, we calculate the aluminum bromide formed from the Al:

5.898 x 10^-3 moles Al (\frac{2 moles AlBr3}{2 moles Al}) (\frac{266.7 g AlBr3}{1 mol AlBr3}) = 1.573 g of Al

5 0
2 years ago
Shaw’s carries two types of apple juice. One is 100% fruit juice, while the other is only 40% juice. Yesterday there was only on
zhuklara [117]

Answer:

the percent of actual fruit juice in the mixture will be 76%

Explanation:

given that we are mixing 2 bottles the total volume the mixture will be

mixture volume = volume bottle 1 + volume bottle 2  = 48 ounces +  32 ounces = 80 ounces

also the total quantity of juice will be

juice mass in mixture = juice bottle 1  + juice bottle 2 = 48 ounces * 1 +  32 ounces * 0.4 = 60.8 ounces of juice

therefore the concentration of juice in the mixture will be

concentration = 60.8 ounces of juice / 80 ounces = 0.76 (76%)

therefore the percent of actual fruit juice in the mixture will be 76%

8 0
2 years ago
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