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Alla [95]
2 years ago
13

How many grams of H3PO4 are in 234 mL of a 2.5 M solution of H3PO4?

Chemistry
1 answer:
motikmotik2 years ago
5 0

Answer:

57 grams of H3PO4

Explanation:

M= moles/ liters

convert mL to L

234 mL x 1L/1000mL = 0.234L

Rearrange the Molarity formula to solve for moles.

moles= MxL

moles= 2.5M x 0.234L

moles= 0.585 mol

Use the molar mass of H3PO4 to get to grams

0.585 mol x 97.994 grams/1 mol = 57.326 grams of H3PO4

round to two sig figs for 57 grams

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Number of H in 3(NH4)2Cro4
zlopas [31]

The number of H atoms in 3(NH₄)₂CrO₄ = 24

<h3>Further explanation  </h3>

The empirical formula is the smallest comparison of atoms of compound forming elements.  

A molecular formula is a formula that shows the number of atomic elements that make up a compound.  

(empirical formula) n = molecular formula  

Subscripts in the chemical formula indicate the number of atoms

The compound of 3(NH₄)₂CrO₄ ( 3 molecules of (NH₄)₂CrO₄ ) :

Number of H :

\tt 4\times 2(subscript)\times 3(coefficient,number~of~molecules)=24~atoms

7 0
3 years ago
If I have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature? (205)
tiny-mole [99]

Answer:

204.73K

Explanation:

the formula : PV=nRT

n=4

P=5.6 atm

V=12 L

R=0.08206 L atm mol-1 K-1

T=?

So, if you plug it in, you will get:-

T=PV/nR

T=(5.6 atm)(12 L)/(4 mol)(0.08206 L atm mol-1 K-1)

T=204.73 K

hope this is correct!

4 0
3 years ago
Some atoms are chemically stable and will not bond with atoms of other
tigry1 [53]

Answer:

The stability of atoms depends on whether or not their outer-most shell is filled with electrons. If the outer shell is filled, the atom is stable. Atoms with unfilled outer shells are unstable, and will usually form chemical bonds with other atoms to achieve stability.

Explanation:

7 0
3 years ago
Read 2 more answers
H3C6H507 + H2O + H3O+ + H2C6H507<br> acid <br> base
Aleonysh [2.5K]

Explanation:

an acid will give away a proton and become a conjugate base.

A base will accept a proton and become a conjugate acid.

3 0
3 years ago
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

R_{in} = \frac{0.04}{100}*2000

R_{in} = 0.8

The rate-out

R_{out} = \frac{A}{6000}*2000

R_{out} = \frac{A}{3}

We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

\frac{dA}{dt} +\frac{A}{3} =0.8

Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

so we have:

e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

7 0
3 years ago
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