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wlad13 [49]
3 years ago
15

If the half life is 6.3 minutes how much of a 128.0 milligram sample will remain after 15 minutes​

Chemistry
1 answer:
zaharov [31]3 years ago
6 0

Answer:

24.5

Explanation:

15/6.3 => 2.38 half lives passed.

.5^2.38 => 0.19198 decimal representation of the percentage that is left over after 2.38 half lives have passed.

0.19198 *128 = 24.5 mg of the material remaining.

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An 80.0-gram sample of water at 10.0°C absorbs 1680 Joules of heat energy. What is the final temperature of the water? a 50.0°C
ICE Princess25 [194]

Answer:

b)15.0°C

Explanation:

Specific Heat of Water=4.2 J/g°C

This means, that 1 g of Water will take 4.2 J of energy to increase its temperature by 1°C.

∴80 g Water will take 80×4.2 J of energy to increase its temperature by 1°C.

80×4.2 J=336 J

Total Energy Provided=1680 J

The temperature increase=\frac{\textrm{Total energy required}}{\textrm{energy required to increase temperature by one degree}}

Temperature increase=\frac{1680}{336}

=5°C

Initial Temperature =10°C

Final Temperature=Initial + Increase in Temperature

=10+5=15°C

7 0
3 years ago
How did Thomson discover the electron?
Hoochie [10]

J. J. Thomson discovered electron by performing an experiment using cathode ray tubes. High voltage across is applied across two electrodes at that  causes a beam of particles to flow from the the negatively-charged electrode that is  cathode to  the positively-charged electrode that is anode. Properties of the particles, are tested using two oppositely-charged electric plates around the cathode ray. The cathode ray was deflected away from the negatively-charged electric plate and towards the positively-charged plate. This indicated that the cathode ray was composed of negatively-charged particles. And these negatively charged particles are called electrons.


4 0
4 years ago
A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If
NemiM [27]

Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

Q_1=-Q_2

Mass of steel= m_1

Specific heat capacity of steel = c_1=0.452 J/g^oC

Initial temperature of the steel = T_1=2.00^oC

Final temperature of the steel = T_2=T=21.50^oC

Q_1=m_1c_1\times (T-T_1)

Mass of water= m_2= 105 g

Specific heat capacity of water=c_2=4.18 J/g^oC

Initial temperature of the water = T_3=22.00^oC

Final temperature of water = T_2=T=21.50^oC

Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))

On substituting all values:

(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}

<h3>25.0 grams is the mass of the steel bar.</h3>
3 0
3 years ago
Read 2 more answers
titration 25.0 ml of 3.4 x 10^ -3 M of Ba(OH) Neutralize 16.6 mL of HCL solution. What is the molarity of HCL solution?
marusya05 [52]
What we  are give: Concentration of base (CB) = 3.4 × 10^{-3}
Then convert all volume in ml to L.
Volume of base (VB) 25.0ml = 0.025L
Volume of acid (VA) 16.6ml = 0.0166L
Now that we have everything we use the formula CAVA=CBVB.
Make 'CA' the subject then solve. 
CA=\frac{CBVB}{VA}

8 0
3 years ago
PLEASE HELP!! due soon
mina [271]

Answer:

The pic won't load sorry try posting this question again and then i'll be able to answer!

Explanation:

4 0
3 years ago
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