PLEASE HELP ITS VERY EASY FOR OTHER PLEASE ITS A TEST DUE IN AN HOUR PLEASEEEEEEEEEE

What is the mass in grams of KBr in 0.400 L of a 0.350 M solution???

melamori03 [73]

Answer:

The answer to your question is: 16.7 g of KBr

Explanation:

Data

mass KBr = ? g

Volume = 0.400 L

Concentration = 0.350 M

Formula

Molarity = moles / volume

moles = molarity x volume

Process

moles = (0.350)(0.400)

= 0.14

MW KBr = 39 + 80 = 119 g

119 g of KBr -------------------- 1 mol

x -------------------- 0.14 mol

x = (0.14 x 119) / 1

x = 16.7 g of KBr

8 0

3 years ago

Read 2 more answers

Please help!!!!! I need the correct answer quickly!!!

xxMikexx [17]

Answer :

The oxidation state of oxygen (O) in $OF_2$ is, (+2)

The oxidation state of carbon (C) in $CO$ is, (+2)

The oxidation state of nitrogen (N) in $K_3N$ is, (-3)

Explanation :

Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of oxygen (O) in compounds is usually -2, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +2.

The oxidation number of a Group 17 element in a binary compound is -1.

The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

(a) The given compound is, $OF_2$

Let the oxidation state of 'O' be, 'x'

$x+2(-1)=0\\\\x-2=0\\\\x=+2$

The oxidation state of oxygen (O) in $OF_2$ is, (+2)

(b) The given compound is, $CO$

Let the oxidation state of 'C' be, 'x'

$x+(-2)=0\\\\x-2=0\\\\x=+2$

The oxidation state of carbon (C) in $CO$ is, (+2)

(c) The given compound is, $K_3N$

Let the oxidation state of 'N' be, 'x'

$3(+1)+x=0\\\\3+x=0\\\\x=-3$

The oxidation state of nitrogen (N) in $K_3N$ is, (-3)

7 0

3 years ago

Read 2 more answers

Consider a transition of the electron in the hydrogen atom from n=5 to n=9. Determine the wavelength of light that is associated

lesantik [10]

Lol i need to answer more and plus n could also = # 4.65

5 0

4 years ago

52.1 mL of aqueous 0.255 M Pb(NO3)2 is mixed with 38.5 mL of 0.415 M NaCl. The equation for the precipitate reaction is: Pb(NO3)

Elan Coil [88]

Answer:

0.0585 M

Explanation:

Pb(NO₃)₂ (aq) + 2NaCl (aq) → PbCl₂ (s) + 2NaNO₃ (aq)

First we calculate the inital number of moles of each reagent, using the given volumes and concentrations:

0.255 M Pb(NO₃)₂ * 52.1 mL = 13.3 mmol Pb(NO₃)₂

0.415 M NaCl * 38.5 mL = 16.0 mmol NaCl

Then we calculate how many Pb(NO₃)₂ moles reacted with 16.0 mmoles of NaCl, using the stoichiometric coefficients of the reaction:

16.0 mmol NaCl * $\frac{1mmolPb(NO_3)_2}{2mmolNaCl}$ = 8.00 mmol Pb(NO₃)₂

Now we calculate the remaining number of Pb(NO₃)₂ moles after the reaction:

13.3 mmol - 8.00 mmol = 5.30 mmol Pb(NO₃)₂

Finally we divide the number of moles by the final volume to calculate the concentration:

5.30 mmol / (52.1 mL + 38.5 mL) = 0.0585 M

6 0

3 years ago

A 6.13 g sample of an unknown salt (MM = 116.82

Olin [163]

Answer:

-3.19x10³ J

Explanation:

Since the surroundings absorbed 3.19 × 10³ J (or 3190 J) of heat, the system, or the dissolution reaction, must have lost the same amount of heat. The heat for the system, then, is -3.19 × 10³ J (or -3190 J). We know this is true because of the first law of thermodynamics, "heat is a form of energy, and thermodynamic processes are therefore subject to the principle of conservation of energy".

6 0

3 years ago