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2 years ago

Identify each of the highlighted materials as an element, a compound, or a mixture, and explain your reasoning.

Chemistry

1 answer:

Reika [66]2 years ago

4 0

Answer:

These are compounds.

Explanation:

As it is formed by chemically bonding of two elements C ( carbon) and H (hydrogen) .

Its not a mixture because mixture is Just dispersed of diferrent elements or a compound which are not in fix ratio

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As the mass of a sample increases, the number of moles present in the sample

vivado [14]

Answer:

increases

Explanation:

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3 years ago

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How many grams of solute are needed to make 37.5 mL of 0.750 M KI solution? Round to three significant digits.

Marysya12 [62]

4.648 gm of solute is needed to make 37.5 mL of 0.750 M KI solution.

Solution:

We will start with the Molarity

$\text { Molarity }=\text { Mole of solute } \div \text { liter of solution }$

Also we know 1000 ml = 1 L

Therefore 37.5 ml by 1000ml we obtained 0.0375L

Equation for solving mole of solute

$\text { Mole of solute }=\text { Molarity } \times \text { Liters of solution }$

Now, multiply 0.750M by 0.0375

Substitute the known values in the above equation we get

$0.750 \times 0.0375=0.0281$

Also we know that Molar mass of KI is 166 g/mol

So divide the molar mass value to get the no of grams.

$0.028 \times 166=4.648$

So 4.648 gm of Solute is required for make 37.5 mL of 0.750 M KI solution.

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3 years ago

Calculate the reaction quotient Qp for the following redox reaction: 14H+ + Cr2O72- + 6Cl- ----> 2Cr3+ + 3Cl2 + 7H2O The reac

stich3 [128]

Answer:

Value of $Q_{p}$ for the given redox reaction is $1.0\times 10^{-8}$

Explanation:

Redox reaction with states of species:

$14H^{+}(aq.)+Cr_{2}O_{7}^{2-}(aq.)+6Cl^{-}(aq.)\rightarrow 2Cr^{3+}(aq.)+3Cl_{2}(g)+7H_{2}O(l)$

Reaction quotient for this redox reaction:

$Q_{p}=\frac{[Cr^{3+}]^{2}.P_{Cl_{2}}^{3}}{[H^{+}]^{14}.[Cr_{2}O_{7}^{2-}].[Cl^{-}]^{6}}$

Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of $H_{2}O$ is taken as 1 due to the fact that $H_{2}O$ is a pure liquid.

$pH=-log[H^{+}]$

So, $[H^{+}]=10^{-pH}$

Plug in all the given values in the equation of $Q_{p}$ :

$Q_{p}=\frac{(0.10)^{2}\times (0.010)^{3}}{(10^{-0.0})^{14}\times (1.0)\times (1.0)^{6}}=1.0\times 10^{-8}$

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