All categories

3 years ago

What is the percentage composition when 10g of magnesium combines with 4g of nitrogen?

1 answer:

7 0

The % composition when 10g of magnesium combine with 4g of nitrogen is 71.43% magnesium and 28.57 % nitrogen

calculation

% composition = mass of an element / total mass x100
mass of magnesium = 10 g
mass of nitrogen = 4g

calculate the total mass used

= 10g of Magnesium + 4 g of nitrogen = 14 grams

% composition for magnesium is therefore = 10/14 x100 = 71.43 %

% composition for nitrogen is therefore = 4 /14 x100 = 28.57 %

You might be interested in

Someone please help fast

Volgvan

Answer:

The last region should be right

Explanation:

3 0

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

5 0

3 years ago

Calculate the molarity of each of the following solutions. Use the periodic table if necessary.

-Dominant- [34]

Answer: 2.8

Explanation: just took the quiz

8 0

2 years ago

Read 2 more answers

A sample of gas occupies 7.80 liters at 425°C? What will be the volume of the gas at 35°C if the pressure does not change?

balandron [24]

From ideal gas equation PV = nRT, V/T = nR/P ==> V/T = constant. Therefore V1/T1 = V2/T2 ==> 7.8/698 = V2/308. V = 3.44L {TEMPERATURE IN KELVIN = 273 + 425 AND 35 = 698 AND 308}

4 0

2 years ago

Zn + I2 ---> Znl2<br> Determine the theoretical yield of the product if 2g of Zn is used

egoroff_w [7]

Answer:

Mass = 9.58 g

Explanation:

Given data:

Mass of Zn = 2g

Theoretical yield of ZnI₂ = ?

Solution:

Chemical equation:

Zn + I₂ → ZnI₂

Number of moles of Zn:

Number of moles = mass/molar mass

Number of moles = 2g / 65.38 g/mol

Number of moles = 0.03 mol

Now we will compare the moles of Zn and ZnI₂.

Zn : ZnI₂

1 : 1

0.03 : 0.03

Mass of ZnI₂:

Mass = number of moles × molar mass

Mass = 0.03 mol × 319.22 g/mol

Mass = 9.58 g

4 0

3 years ago