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2 years ago

If you have 45.6mL of a liquid and its

Chemistry

1 answer:

tankabanditka [31]2 years ago

4 0

The density of the liquid is 1.027g/cm³

D=M/V
D=46.83/45.6
D=1.027g/cm³

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Jenny places a strip of pH paper into a solution. When she removes the pH paper, it has turned yellow-green.

Ronch [10]

Jenny puts the ph paper and lines it up PH SCALE to find how strong the solution.

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3 years ago

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How many grams of CaCl2 are needed to make 277.8g of a solution that is 31.5% (m/m) in water? Note that mass is not technically

Sveta_85 [38]

Answer: The mass of calcium chloride present in given amount of solution is 87.5 g

Explanation:

We are given:

Mass of solution = 277.8 grams

Also, 31.5 % (m/m) of calcium chloride in water. This means that 31.5 g of calcium chloride is present in 100 g of solution.

To calculate the mass of calcium chloride in the given amount of solution, we use unitary method:

in 100 g of solution, the mass of calcium chloride present is 31.5 g

So, 277.8 g of solution, the mass of calcium chloride present is $\frac{31.5}{100}\times 277.8=87.5g$

Hence, the mass of calcium chloride present in given amount of solution is 87.5 g

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3 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li

meriva

Answer:

The standard enthalpy change for the reaction at $25^{0}\textrm{C}$ is -2043.999kJ

Explanation:

Standard enthalpy change ( $\Delta H_{rxn}^{0}$ ) for the given reaction is expressed as:

$\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]$

Where $\Delta H_{f}^{0}$ refers standard enthalpy of formation

Plug in all the given values from literature in the above equation:

$\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ$

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2 years ago

Which of the following elements will form a positive ions? check all that apply

tamaranim1 [39]

we're the rest of the question

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2 years ago

How much of nacl is in 1.67 l of 0.400 m nacl? answer in units of mol.

Evgen [1.6K]

Answer:

.668 mole

Explanation:

1.67 * .4 = .668 mole

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1 year ago