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3 years ago

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Hydrogen bonds within liquid water are attractions between protons and hydroxide ions. are dipole-dipole attractions. are ion-in

duced dipole attractions. are attractions between protons and oxygen nuclei. are attractions between two hydrogen atoms.

1 answer:

barxatty [35]3 years ago

6 0

Answer:

true because the bonds cannot be broken down

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Carbon reacts with oxygen to produce carbon dioxide (CO2).

Jobisdone [24]

2 different elements- carbon and oxygen.............................

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3 years ago

A 25.0 ml solution of .0600 M EDTA was added to a 57.0 ml sample containing an unknown concentration of V3+. All V3+ present for

Rom4ik [11]

Explanation:

Let us assume that the ratio for the given reaction is 1:1.

Therefore, we will calculate the moles of $Ga^{3+}$ as follows.

Moles of $Ga^{3+}$ solution = molarity × volume (L)

= 0.0440 M × 0.014 L

= 0.000616 moles

Moles of excess EDTA = 0.000616 moles

Also, the initial moles of EDTA will be calculated as follows.

Total initial moles of EDTA = 0.0600 M × 0.025 L

= 0.0015

Therefore, moles of EDTA reacted with $V^{3+}$ will be as follows.

= 0.0015 - 0.000616

= 0.00088 moles

Since, we have supposed a 1 : 1 ratio between $V^{3+}$ and EDTA .

So, moles of $V^{3+}$ = 0.00088 moles

Now, we will calculate the molarity of $V^{3+}$ as follows.

Molarity of $V^{3+}$ solution = $\frac{\text{moles}}{\text{volume (L)}}$

= $\frac{0.00088 moles}{0.057 L}$

= 0.015 M

Thus, we can conclude that the original concentration of the $V^{3+}$ solution is 0.015 M.

4 0

3 years ago

The Questions are in the Pictures Attached

Whitepunk [10]

Answer:

1 is 5<3=me & you

Explanation:

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3 years ago

How many moles of O2 react with .75 moles of c10h22

kap26 [50]

Answer:

2 moles

Explanation:

7 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

4 years ago