Calculate the molar mass

If D+2 would react with E-1, what do you predict to be the formula?

harkovskaia [24]

Answer:

DE2

Explanation: for every one D+2 you need two E-1 because +2=-2

8 0

2 years ago

Jarvis checked his odometer reading when filling his tank, and then checked it several days later when he filled his tank again.

8_murik_8 [283]

Answer: 26.25 miles per gallon

Explanation: first odometer reading = 65290

Second odometer reading = 65500

Difference in odometer reading = second reading - first reading

= 65500 - 65290 = 210 miles.

If it took 8 gallons to fill his tank, it means

8 gallons will get him 210 miles, therefore 1 gallon will get 210 miles/8gallons = 26.25miles/gallon. Hence he is getting 26.25miles/gallon.

4 0

2 years ago

Read 2 more answers

For each pair of bonds, indicate the more polar bond, and use an arrow to show the direction of polarity in each bond.a. C-O and

Naddika [18.5K]

Answer:

The more polar bond is C-O.

Explanation:

The greater polarity is due to that the Oxygen atom is more electronegative than the Nitrogen atom, so the negativity of the molecule tends to be on that side.

7 0

8 months 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}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

5 0

2 years ago

The types of nuclear reactions are fission , fusion , what else

Mrrafil [7]

And also explosion tha is another type

3 0

2 years ago