At STP, which element is brittle and not a conductor of electricity

Chemistry

1 answer:

Sever21 [200]3 years ago

5 0

Your answer is the element of S I think.

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To what volume should you dilute 50.0 ml of 12 m hno3 solution to obtain a 0.100 m hno3 solution?

bearhunter [10]

Answer:

The answer is "6L"

Explanation:

Formula:

$\bold{C_1 \times V_1 = C_2 \times V_2 }\\\\V_2= \frac{C_1\times V_1}{C_2}$

Values:

$\to C_1= 12 \ m\\\to V_1= 50 \ ml\\\to C_2= 0.100 \ m\\\\\\V_2= \frac{12 \times 50 }{0.100}$

$= \frac{12 \times 50 }{0.100}\\\\= \frac{12 \times 50 \times 1000}{100}\\\\= \frac{600 \times 1000}{100}\\\\= 600 \times 10\\\\=6000 \ ml\\= 6 \ L$

4 0

3 years ago

A solution of NaOH is titrated with H2SO4. It is found that 20.05 mL of 0.3564 M H2SO4 solution is equivalent to 43.42 mL of NaO

Darya [45]

Answer : The concentration of NaOH is, 0.336 M

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=0.3564M\\V_1=20.05mL\\n_2=1\\M_2=?\\V_2=43.42mL$

Putting values in above equation, we get:

$2\times 0.3564M\times 20.05mL=1\times M_2\times 43.42mL$

$M_2=0.336M$

Thus, the concentration of NaOH is, 0.336 M

3 0

3 years ago

What is the mass in grams of a single formula unit of silver chloride, AgCI? A) 4.21 x 1021 g B) 8.61 x 10258 C) 1.66 x 10-248 D

bekas [8.4K]

Answer: D) $2.38*10^-^2^2$ g

Explanation: The question asks to convert formula unit to grams. It is a unit conversion problem.

1 mole equals to Avogadro number of formula units. So, to convert the given number of formula units to moles, we need to divide by the Avogadro number. After this, we do moles to grams conversion and for this the moles are multiplied by the molar mass of the compound. Molar mass of AgCl is 143.32 gram per mol.

$1FormulaUnitAgCl(\frac{1mol}{6.022*10^2^3formulaUnits})(\frac{143.32g}{1mol})$