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seropon [69]
3 years ago
6

If an effusion experiment, it required 45 seconds for

Chemistry
1 answer:
patriot [66]3 years ago
8 0

Answer: A  vacuum. Under the same conditions 16 s were required for the same number of moles of O2 to effuse. ... The molar mass of the unknown gas is 200 g/mol. calculation ... QUESTION 1 If gas-burning appliances are vented into an unlined masonry chimney, deteriorate and crumble and may · PLS HELP

Explanation:

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You are a marine biologist who wants to study what type of zooplankton (microscopic animals that live in the ocean and feed larg
timama [110]

Answer:

The advantage of the compound light microscope over the dissecting microscope is the magnification power of the telescope. Compound light microscope magnifies from 40x up to 1,000x while dissecting microscope magnifies up to 40 x only. In this regard, more magnification power is advantageous to view smaller objects. 

4 0
3 years ago
Use the solubility rules to determine if:
xenn [34]

<u>Answer:</u>

A reaction is said to occur if there is a formation of an insoluble solid or a precipitate(s) or a liquid (l) or a gas(g).

If both the reactants and products are in aqueous state, No reaction takes place.

$2 \mathrm{NaCl}(a q)+M g B r_{2}(a q)>M g C l_{2}(a q)+2 \mathrm{NaBr}(a q)(\mathrm{NO} \text { REACTION })$

All chlorides and Bromides are soluble except that of Ag, Hg and Pb.

Hence, No reaction takes place since all the reactants and products are in aqueous states.

${KOH}(a q)+{NaCl}(a q)>K C l(a q)+{NaOH}(a q)(\mathrm{NO} \text { REACTION })$

Salts  of Group IA are soluble. Hence No reaction takes place

$M g S(a q)+2 {NaOH}(a q)>M g(O H)_{2}(s)+N a_{2} S(a q)$

(REACTION TAKES PLACE)

All hydroxides are insoluble except that of Group IA, ammonium ion and  Group IIA down from Calcium.

Hence Reaction takes place with the formation of Mg(OH)_2 precipitate

5 0
3 years ago
Combustion of 9.511 grams of c4h10 will yield ____ grams of CO2
Flauer [41]

Answer:

\boxed{28.81}

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:      58.12                   44.01

           2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

m/g:     9.511

1. Moles of C₄H₁₀

\text{Moles of C$_{4}$H$_{10} $} = \text{ 9.511 g C$_{4}$H$_{10} $} \times \dfrac{\text{1 mol C$_{4}$H$_{10} $}}{\text{ 58.12 g C$_{4}$H$_{10} $}} = \text{0.1636 mol C$_{4}$H$_{10}$}

2. Moles of CO₂

The molar ratio is 8 mol CO₂:2 mol C₄H₁₀

\text{Moles of CO}_{2} =\text{0.1636 mol C$_{4}$H$_{10} $} \times \dfrac{\text{8 mol CO}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \text{0.6546 mol CO}_{2}

3. Mass of CO₂

\text{Mass of CO}_{2} = \text{0.6546 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO}_{2}} = \textbf{28.81 g CO}_{2}\\\\\text{The combustion will form $\boxed{\textbf{28.81 g CO}_{2}}$}

8 0
3 years ago
A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is −0.3 kJ/K
Karolina [17]

Answer:

W = -120 KJ

Explanation:

Since the piston–cylinder assembly undergoes an isothermal process, then the temperature is constant.

Thus; T1 = T2 = 400K

change in entropy; ΔS = −0.3 kJ/K

Formula for change in entropy is written as;

ΔS = Q/T

Where Q is amount of heat transferred.

Thus;

Q = ΔS × T

Q = -0.3 × 400

Q = -120 KJ

From the first law of thermodynamics, we can find the workdone from;

Q = ΔU + W

Where;

ΔU is Change in the internal energy

W = Work done

Now, since it's an ideal gas model, the change in internal energy is expressed as;

ΔU = m•C_v•ΔT

Where;

m is mass

C_v is heat capacity at constant volume

ΔT is change in temperature

Now, since it's an isothermal process where temperature is constant, then;

ΔT = T2 - T1 = 0

Thus;

ΔU = m•C_v•ΔT = 0

ΔU = 0

From earlier;

Q = ΔU + W

Thus;

-120 = 0+ W

W = -120 KJ

8 0
3 years ago
Astatine-210 has a half-life of 8.08 days. What fraction of a sample of astatine-210 is left unchanged after 16.16 days?
Flauer [41]

Answer:

0

Explanation:

Given parameters:

Half-life  = 8.08days

Unknown:

What fraction is left unchanged after 16.16days = ?

Solution:

The half - life of a substance is the time taken for the half of a radioactive material to decay to half.

 

  Day 0           Day 8.08         Day 16.16

   100%                 50%                 0%        Parent

    0%                     50%                100%     Daughter

After 16.16 days, non of the original sample will remain unchanged.

6 0
3 years ago
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