Ask question

Ask question

All categories

3 years ago

6

You have 200 g of a substance with a molar mass of 150 g/mol. How many moles of the substance do you have? 0.75 mol 1.00 mol 1.3

3 mol 1.50 mol

1 answer:

tigry1 [53]3 years ago

7 0

Answer:

Explanation:

200 g/150 g/mol = 1.33 mol

You might be interested in

-5 - -2 plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz helppppppppppppppppppppppppppppppp meee

Zepler [3.9K]

The answer would be -3

4 0

3 years ago

Read 2 more answers

__CH4+__CL2> __CCL4+__HCL

tatyana61 [14]

Answer:

CH4+4Cl2->CCl4+4HCl

6 0

3 years ago

When 100. mL of 6.00 M HCl is diluted to 300. mL, the final concentration is ________.

Marat540 [252]

Answer:

2.00 M

Explanation:

The concentration of a solution is given by

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

where

m is the mass of solute

V is the volume of the solution

At the beginning, the solution has:

M = 6.00 M is the concentration

V = 100 mL = 0.1 L is the volume

So the mass of solute (HCl) is

$m=MV=(6.00)(0.1)=0.6g$

Then, the HCL is diluted into a solution with volume of

V = 300 mL = 0.3 L

Therefore, the final concentration is:

$M=\frac{m}{V}=\frac{0.6}{0.3}=2.00 M$

3 0

4 years ago

Read 2 more answers

A chemist dissolves 484 .mg of pure perchloric acid in enough water to make up 240.mL of solution. Calculate the pH of the solut

sdas [7]

Answer:

1.70

Explanation:

The molar mass of perchloric acid is 100.46 g/mol. The moles corresponding to 484 mg (0.484 g) are:

0.484 g × (1 mol/100.46 g) = 4.82 × 10⁻³ mol

4.82 × 10⁻³ moles are dissolved in 240 mL (0.240 L) of solution. The molar concentration of perchloric acid is:

4.82 × 10⁻³ mol/0.240 L = 0.0201 M

Perchloric acid is a strong monoprotic acid, that is, it dissociates completely, so [H⁺] = 0.0201 M.

The pH is:

pH = -log [H⁺] = -log 0.0201 = 1.70

8 0

3 years ago

If 50 g of lead (of specific heat 0.11 kcal/kg ∙ C°) at 100°C is put into 75 g of water (of specific heat 1.0 kcal/kg ∙ C°) at 0

Contact [7]

Answer: The final temperature is $279.8K$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of lead = 50 g

$m_2$ = mass of water = 75 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of lead = $100^oC=373K$

$T_2$ = temperature of water = $0^oC=273K$

$c_1$ = specific heat of lead = $0.11kcal/kg^0C$

$c_2$ = specific heat of water= $1.0kcal/kg^0C$

Now put all the given values in equation (1), we get

$50\times 0.11\times (T_{final}-373)=-[75\times 1.0\times (T_{final}-273)]$

$T_{final}=279.8K$

Therefore, the final temperature of the mixture will be 279.8 K.

3 0

4 years ago