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3 years ago

Pls help me no links

Chemistry

2 answers:

uysha [10]3 years ago

8 0

If I remember correctly it should be mixture

Assoli18 [71]3 years ago

6 0

Answer:

mixture

Explanation:

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True or false. for all atoms of the same element the 4s orbital is larger than the 3s orbital

Alenkinab [10]

Answer:

Between 4s and 3s orbital , 3s has more energy .

Explanation:

According to the rule , the lower the value of (n+l) for an orbital , the lower is it's energy . And if two orbitals have the same value of (n+l), the orbital with lower value of n will have the lower energy .

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3 years ago

The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.

olchik [2.2K]

Answer:

Approximately $0.291\; \rm M$ (rounded to two significant figures.)

Explanation:

The unit of concentration $\rm M$ is the same as $\rm mol \cdot L^{-1}$ (moles per liter.) On the other hand, the volume of both the $\rm NaOH$ solution and the original $\rm HCl$ solution here are in milliliters. Convert these two volumes to liters:

$V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L$ .
$V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L$ .

Calculate the number of moles of $\rm NaOH$ in that $0.0182\; \rm L$ of $0.160\; \rm M$ solution:

$\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}$ .

$\rm HCl$ reacts with $\rm NaOH$ at a one-to-one ratio:

$\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)$ .

Coefficient ratio:

$\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1$ .

In other words, one mole of $\rm NaOH$ would neutralize exactly one mole of $\rm HCl$ . In this titration, $0.291\; \rm mol$ of $\rm NaOH\!$ was required. Therefore, the same amount of $\rm HC$ should be present in the original solution:

$\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}$ .

Calculate the concentration of the original $\rm HCl$ solution:

$\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

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3 years ago

What is the third ionization energy of titanium?

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Ti^2+(g)-->Ti^3+(g)+-3rd IP=2652.5

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3 years ago

Which of the following provides evidence that supports the continental drift theory? PLEASE HELP 10 POINTS

Tanya [424]

I believe that it is A. If you remember in lesson 03.03 it gave multiple examples of things that support the continental drift theory. One of them was fossils of the same organisms found in different continents.

4 0

3 years ago

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Use the equation to determine what mass of FeS must react to form 326 g of FeCl2.

iogann1982 [59]

Answer:

We need 226 grams of FeS

Explanation:

Step 1: Data given

Mass of FeCl2 = 326 grams

Molar mass FeCl2 = 126.75 g/mol

Step 2: The balanced equation

FeS + 2 HCl → H2S + FeCl2

Step 3: Calculate moles FeCl2

Moles FeCl2 = 326 grams / 126.75 grams

Moles FeCl2 = 2.57 moles

Step 4: Calculate moles FeS needed

For 1 mol H2S and 1 mol FeCl2 produced, we need 1 mol FeS and 2 moles HCl

For 2.57 moles FeCl2 we need 2.57 moles FeS

Step 5: Calculate mass FeS

Mass FeS = 2.57 moles * 87.92 g/mol

Mass FeS = 226 grams FeS

We need 226 grams of FeS

4 0

3 years ago