Using the periodic table, find the neutral atom that has the same electron configuration as 1s²2s22p.

What is the mass of bismuth carbonate (597.99 g/mol) that decomposes to release 50.0ml of carbon dioxide gas at stp? 0.000445 g

REY [17]

The mass of bismuth carbonate that decomposed is 0.445 g

calculation

write the equation for reaction

that is Bi2(CO3)3 → Bi2O3 +3 CO2

find the moles of of CO2 formed

At STP 1 mole of a gas = 22.4 L

what about 50 Ml

convert the Ml to Liters = 50/1000= 0.05 L

moles of CO2 formed is therefore= 1 mole x 0.05 L/22.4 l =2.232 x10^-3 moles

by use of reacting ratio between Bi(CO3)3 to CO2 which is 1:3 the moles of Bi(CO3)3 is = 2.232 x10^-3 x1/3= 0.744 x10^-4 moles of Bi(CO3)3

mass of Bi(CO3)3 = molar mass x number of moles

mass= 0.744 x10^-4 moles x 597.99 g/mol = 0.445 grams

5 0

2 years ago

In addition to carbon, hydrogen and oxygen, amino acids also contain?

N76 [4]

Amino acids also contain nitrogen

6 0

3 years ago

Please help me i don’t understand.

vekshin1

Answer:

Your answer is

$p = m \div v \\ = \: 38 \: gram \: \div 2 \: cm \\ = 19 \: gram \: = 19000gm$

Hope it helped

8 0

3 years ago

A solution contains 42.0 g of heptane (C7H16) and 50.5 g of octane (C8H18) at 25 ∘C. The vapor pressures of pure heptane and pur

Kruka [31]

Answer:

(a) 22.3 torr; 5.6 torr; (b) 27.9 torr; (c) 77.7 % heptane; 23.3 % octane

(d) Heptane is more volatile than octane

Explanation:

We can use Raoult's Law to solve this problem.

It states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction. In symbols,

$p_{i} = \chi_{i} p_{i}^{\circ}$

(a) Vapour pressure of each component

Let heptane be Component 1 and octane be Component 2.

(i) Moles of each component

$n_{1} = \text{42.0 g} \times \dfrac{\text{1 mol}}{\text{100.20 g}} = \text{0.4192 mol}\\n_{2} = \text{50.5 g} \times \dfrac{\text{1 mol}}{\text{114.23 g}} = \text{0.4421 mol}$

(ii) Total moles

$n_{\text{tot}} = 0.4192 + 0.4421 = \text{0.8613 mol}$

(iiii) Mole fractions of each component

$p_{1} = 0.4867 \times 45.8 = \textbf{22.3 torr}\\p_{2} = 0.5133 \times 10.9 = \ \textbf{5.6 torr}$

(iv) Partial vapour pressures of each component

$p_{1} = 0.4867 \times 45.8 = \textbf{22.3 torr}\\p_{2} = 0.5133 \times 10.9 = \textbf{5.6 torr}$

(b) Total pressure

$p_{\text{tot}} = p_{1} + p_{2} = 22.3 + 5.6 = \text{27.9 torr}$

(c) Mass percent of each component in vapour

$\chi_{1} = \dfrac{p_{1}}{p_{\text{Tot}}} = \dfrac{22.3}{27.9} =0.799\\\chi_{2} = \dfrac{p_{2}}{p_{\text{Tot}} }= \dfrac{5.6}{27. 9} =0.201$

The ratio of the mole fractions is the same as the ratio of the moles.

$\dfrac{n_{1}}{n_{2}} = \dfrac{0.799}{0.201}$

If we have 1 mol of vapour, we have 0.799 mol of heptane and 0.201 mol of octane

$m_{1} = 0.799 \times 100.20 = \text{80.1 g}\\m_{2} = 0.201\times 114.23 = \text{23.0 g}\\m_{\text{tot}} = 80.1 + 23.0 = \text{103.1 g}\\\\\text{ mass percent heptane} = \dfrac{80.1}{103.1} \times 100 \, \% = \mathbf{77.7\, \%}\\\\\text{ mass percent octane} = \dfrac{23.0}{103.1} \times 100 \, \% = \mathbf{22.3\, \%}}$

(d) Enrichment of vapour

The vapour is enriched in heptane because heptane is more volatile than octane.

5 0

3 years ago

Please help me with this question

Tcecarenko [31]

Answer:

Moon

Explanation:

5 0

2 years ago