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Mkey [24]
3 years ago
8

sample of saline contains 45 ml of saline in 128 ml of water. What is the percent by volume of the solution?

Chemistry
1 answer:
lana66690 [7]3 years ago
4 0
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What volume of a 2.0M NaOH(aq) is needed to completely neutralize 24 milliliters of 0.5M HCl(aq)?
borishaifa [10]

Answer:

V_{base}=6.0mL

Explanation:

Hello there!

In this case, by considering that the reaction between sodium hydroxide and hydrochloric acid is in a 1:1 mole ratio of these two reactants, we are able to use the following equation relating the concentration and volume of each one:

M_{acid}V_{acid}=M_{base}V_{base}

In such a way, by solving for the volume of the base, we will obtain:

V_{base}=\frac{M_{acid}V_{acid}}{M_{base}} \\\\V_{base}=\frac{0.5M*24mL}{2.0M}\\\\V_{base}=6.0mL

Regards!

8 0
3 years ago
Match the following chemical reactions:
S_A_V [24]

Answer:

Synthesis - 4

reversible- 2

exchange- 1

decomposition-3

Explanation:

In synthesis reaction two or more components combines to form a single product. example 2H2+O2⇒2H2O

In reversible reaction two reactants combine to form two products . The products then reacts and forms back the reactants. example N2 +3H2 ⇒2NH3

In exchange reaction there is an alternation of ions of reactants to form new products. AB+CD ⇒AC + BD

In decomposition reaction, molecules of a compound break down by the action of heat or light or catalyst. example CaCO3 ⇒CaO +CO2

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
Which is the molar mass of H2O?
Svetllana [295]
H=(2x1.008)=2.016
O= 15.999

15.999
+ 2.016
______
18.015 g/mol :)

3 0
3 years ago
Read 2 more answers
HELP PLEASE!! ASAP What would the mass be in kg?
fgiga [73]

Answer:

i guess answer is 0.0600

Explanation:

Here,

Density=0.791g/cm^3

Volume=75.89ml

So,

Mass=Density*Volume

=0.791g/cm^3×75.89ml

=60.02g/cm^3.ml

Expressing them in kg

60.02/1000

=0.0600 kg

4 0
3 years ago
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