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2 years ago

What do the noble gases have in common with the halogens? Both noble gases and halogens are

1 answer:

5 0

The halogens and noble gases are alike because they are all non-metallic elements, with the exception of astatine, a radioactive

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Which of these is NOT a physical change?

alexdok [17]

Answer:

C. Magnetization of iron

Explanation:

This is a chemical change and cannot be undone by physical means therefore it isn't a physical change

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3 years ago

If you created a food chain using the following items. which item would follow the corn?

liq [111]

1. chicken
2. corn
3. human
4. A food chain shows the flow of energy from one item to the next but a food web does not.
a food chain is a small piece of a larger food web. Is true

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3 years ago

Read 2 more answers

Red giant stars defination

kramer

Answer:

A red giant is a luminous giant star of low or intermediate mass in a late phase of stellar evolution. The outer atmosphere is inflated and tenuous, making the radius large and the surface temperature around 5,000 K or lower.

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2 years ago

Read 2 more answers

Determine the molarity for each of the following solution solutions:

____ [38]

Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c)The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d)The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e)The molarity of $Br_2$ solution is, 0.0565 mole/L

(f)The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

Explanation :

(a) 1.457 mol of KCl in 1.500 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Solute is KCl.

$\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L$

The molarity of KCl solution is, 0.9713 mole/L

(b) 0.515 gram of $H_2SO_4$ , in 1.00 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $H_2SO_4$

Molar mass of $H_2SO_4$ = 98 g/mole

$\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L$

The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c) 20.54 g of $Al(NO_3)_3$ in 1575 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $Al(NO_3)_3$

Molar mass of $Al(NO_3)_3$ = 213 g/mole

$\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L$

The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d) 2.76 kg of $CuSO_4.5H_2O$ in 1.45 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $CuSO_4.5H_2O$

Molar mass of $CuSO_4.5H_2O$ = 250 g/mole

$\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L$

The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e) 0.005653 mol of $Br_2$ in 10.00 ml of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Solute is $Br_2$ .

$\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L$

The molarity of $Br_2$ solution is, 0.0565 mole/L

(f) 0.000889 g of glycine, $C_2H_5NO_2$ , in 1.05 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $C_2H_5NO_2$

Molar mass of $C_2H_5NO_2$ = 75 g/mole

$\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L$

The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

5 0

3 years ago

The empirical formula of a compound is CH. At 200 degree C, 0.145 g of this compound occupies 97.2 mL at a pressure of 0.74 atm.

Vladimir [108]

Answer:

The molecular formula = $C_{6}H_{6}$

Explanation:

Given that:

Mass of compound, m = 0.145 g

Temperature = 200 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (200 + 273.15) K = 473.15 K

V = 97.2 mL = 0.0972 L

Pressure = 0.74 atm

Considering,

$n=\frac{m}{M}$

Using ideal gas equation as:

$PV=\frac{m}{M}RT$

where,

P is the pressure

V is the volume

m is the mass of the gas

M is the molar mass of the gas

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the values in the above equation as:-

$0.74\times 0.0972=\frac{0.145}{M}\times 0.0821\times 473.15$

$M=78.31\ g/mol$

The empirical formula is = $CH$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12 + 1 = 13 g/mol

Molar mass = 78.31 g/mol

So,

Molecular mass = n × Empirical mass

78.31 = n × 13

⇒ n ≅ 6

The molecular formula = $C_{6}H_{6}$

6 0

3 years ago