Please do this please please it depends on my grades

The flamability (burns) is a physical change ?

NikAS [45]

Flammability is a chemical change because when you burn something, it no longer has the same properties.

5 0

3 years ago

Read 2 more answers

What is the predicted change in the boiling point of water when 1.50 g of

dezoksy [38]

Answer:

0.00735°C

Explanation:

By seeing the question, we can see the elevation in boiling point with addition of BaCl₂ in water

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$\textsf {While} \: \sf {\Delta T_b} \: \textsf{expression is used} \\ \textsf {for elevation of boiling point}$

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The elevation in boiling point is a phenomenon in which there is increase in boiling point in solution, when the particular type of solute is added to pure solvent.

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$\sf \large \underline{The \: formula \: to \: be \: used \: in \: this \: question \: is} \\ \boxed{T_b = i \times K_b \times m}$

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Where 'i' is van't hoff factor which represents the ratio of observed osmotic pressure and the value to be expected.

and 'i' is 3 (as given in the question)

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'Kb' is molal boiling point constant. And it's value is 0.51°C/mol(given in question)

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'm' represent the molality of solution. Molatity is no. of moles of solution present in 1kg of solution.

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To find molality, we have to divide no. of moles of solute by weight of solution

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While first we need to no. of moles

$\sf \implies no. \: of \: moles = \frac{weight \: of \: solute}{molar \: mass \: of \: solute} \\ \\ \implies \sf no. \: of \: moles = \frac{1.5}{208.23} \\ \\ \sf \implies no. \: of \: moles = 0.0072$

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Now, we will find molality

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$\sf \hookrightarrow molality = \frac{no.\: of \: moles}{weight \: of \: solution} \\ \\ \sf \hookrightarrow molality = \frac{0.072}{1.5} \\ \\ \sf \hookrightarrow molality = 0.048 \: mol {kg}^{ - 1}$

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$\textsf{ \large{ \underline{Now substituting the required values}}}$

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$\sf \longmapsto \Delta T_b = 3 \times 0.51 \times 0.0048 \\ \\ \\ \boxed{ \tt{ \longmapsto \Delta T_b =0.00735{ \degree}C}}$

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Henceforth, the change in boiling point is 0.00735°C.

7 0

1 year ago

When 3.00 g of sulfur are combined with 3.00 g of oxygen, 6.00 g of sulfur dioxide (SO2) are formed. What mass of oxygen would b

Drupady [299]

Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.

2 S + 3 O₂ → 2 SO₃

The stoichiometric calculations is as follows:

7 g S * 1 mol/32.06 g S = 0.218 mol S
Moles O₂ needed = 0.218 mol S * 3 mol O₂/2 mol S = 0.3275 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.3275 mol O₂ * 32 g/mol = 10.48 g O₂

3 0

3 years ago

Of the elements fe li te u and he which are considered group b elements

VLD [36.1K]

A group B element is Fe

4 0

3 years ago

in the following reaction, how many grams of benzene (C6H6) will produce 42 grams of CO2? 2C6H6 + 15O2 → 12CO2 + 6H2O

Mrrafil [7]

Answer: -

12.41 g

Explanation: -

Mass of CO₂ = 42 g

Molar mass of CO₂ = 12 x 1 + 16 x 2 = 44 g / mol

Number of moles of CO₂ = $\frac{42}{44 g/mol}$

= 0.9545 mol

The balanced chemical equation for this process is

2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

From the balanced chemical equation we see

12 mol of CO₂ is produced from 2 mol of C₆H₆

0.9545 mol of CO₂ is produced from $\frac{2 mol C6H6 x 0.9545 mol CO2}{12 mol CO2}$

= 0.159 mol of C₆H₆

Molar mass of C₆H₆ = 12 x 6 + 1 x 6 =78 g /mol

Mass of C₆H₆ =Molar mass x Number of moles

= 78 g / mol x 0.159 mol

= 12.41 g

8 0

3 years ago