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3 years ago

Which of the following statements about metalloids is true?

Chemistry

2 answers:

Klio2033 [76]3 years ago

7 0

Metalloids had properties that fall between those of metals and nonmetals (I believe that to be correct-.-)

aleksklad [387]3 years ago

3 0

Answer: Option (c) is the correct answer.

Explanation:

Metalloids are the substances or elements that show properties of both metals and non-metals.

Metals are good conductors of heat and electricity. They have shiny surface and can be drawn into thin wires and sheets.

Whereas non-metals are bad conductors of heat and electricity. They are brittle in nature and are not shiny.

Similarly, metalloids are shiny but they are brittle at the same time with a relatively good electrical conductivity.

Thus, we can conclude that the statement, metalloids have properties that fall between those of metals and nonmetals, is true about metalloids.

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A metal sample weighing 129.00 grams and at a temperature of 97.8 degrees Celsius was placed in 45.00 grams of water in a calori

Nitella [24]

Answer : The specific heat of metal is $0.481J/g^oC$ .

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of metal = 129.00 g

$m_2$ = mass of water = 45.00 g

$T_f$ = final temperature = $39.6^oC$

$T_1$ = initial temperature of metal = $97.8^oC$

$T_2$ = initial temperature of water = $20.4^oC$

Now put all the given values in the above formula, we get

$129.00g\times c_1\times (39.6-97.8)^oC=-45.00g\times 4.184J/g^oC\times (39.6-20.4)^oC$

$c_1=0.481J/g^oC$

Therefore, the specific heat of metal is $0.481J/g^oC$ .

4 0

3 years ago

What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL o

Nina [5.8K]

Answer:

The concentration of NaCl = 0.3374 M

Explanation:

Given :

Molarity of AgNO₃ = 0.2503 M

Volume of AgNO₃ = 20.22 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 20.22×10⁻³ L

Molarity of a solution is the number of moles of solute present in 1 L of the solution.

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

The formula can be written for the calculation of moles as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Thus,

$Moles\ of\ AgNO_3 =Molarity \times {Volume\ of\ the\ solution}$

$Moles\ of\ AgNO_3 =0.2503 \times {20.22\times 10^{-3}}\ moles$

$Moles\ of\ AgNO_3 = 5.0611 \times 10^{-3} moles$

The chemical reaction taking place:

$AgNO_3_(aq) + NaCl_(aq) \rightarrow AgCl_(s) + NaNO_3_(aq)$

According to reaction stoichiometry:

1 mole of AgNO₃ reacts with 1 mole of NaCl

Thus,

5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl

Thus, moles of NaCl required = 5.0611×10⁻³ moles

Volume of NaCl required = 15.00 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 15.00×10⁻³ L

Applying in the formula of molarity as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity\ of\ NaCl=\frac{5.0611\times 10^{-3}}{15.00\times 10^{-3}}$

$Molarity\ of\ NaCl= 0.3374 M$

Thus, the concentration of NaCl = 0.3374 M

6 0

3 years ago

What volume of a 2.5 M stock solution of acetic acid (HC2H3O2) is required to prepare

Schach [20]

Answer is: volume is 20 mL.
c₁(CH₃COOH) = 2,5 M.
c₂(CH₃COOH) = 0,5 M.
V₂(CH₃COOH) = 100 mL.
V₁(CH₃COOH) = ?
c₁(CH₃COOH) · V₁(CH₃COOH) = c₂(CH₃COOH) · V₂(CH₃COOH).
2,5 M · V₁(CH₃COOH) = 0,5 M · 100 mL.
V₁(CH₃COOH) = 0,5 M · 100 mL ÷ 2,5 M.
V₁(CH₃COOH) = 20 mL ÷ 1000 mL/L =0,02 L.

8 0

3 years ago

An oxide of aluminum contains 0.545 g of Al and 0.485 g of O. Find the empirical formula

uysha [10]

Answer:

The answer is: Al2O3

Explanation:

The data they give us is:

0.545 gr Al

0.485 gr O.

To find the empirical formula without knowing the grams of the compound, we find it per mole:

0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al

0.485 g O * 1 mol O / 16 g O = 0.03 mol O

Then we must divide the results obtained by the lowest result, which in this case is 0.02:

0.02 mol Al / 0.02 = 1 Al

0.03 mol O / 0.02 = 1.5 O

Since both numbers have to give an integer, multiply by 2 until both remain integers:

1Al * 2 = 2Al

1.5O * 2 = 3O

Now the answer is given correctly:

Al2O3

8 0

3 years ago

A sample of CO2 weighing 86.34g contains how many molecules?

irakobra [83]

Answer:

1.181 × 10²⁴ molecules CO₂

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

86.34 g CO₂

Step 2: Identify Conversion

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

Step 3: Convert

$86.34 \ g \ CO_2(\frac{1 \ mol \ CO_2}{44.01 \ g \ CO_2} )(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2} )$ = 1.18141 × 10²⁴ molecules CO₂

Step 4: Check

We are given 4 sig figs. Follow sig fig rules and round.

1.18141 × 10²⁴ molecules CO₂ ≈ 1.181 × 10²⁴ molecules CO₂

4 0

3 years ago