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KATRIN_1 [288]
3 years ago
9

The answer to (3.540)x(0.0065)x(401) should have ____.

Chemistry
1 answer:
miv72 [106K]3 years ago
3 0
B. Two significant figures


Your final answer can only have as many significant figures as the number that has the least amount of significant figures in it, which is (0.0065)
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0.415 g of an unknown triprotic acid are used to make a 100.00 mL solution. Then 25.00 mL of this solution is transferred to an
Arturiano [62]

Answer:

Explanation:

Initial burette reading = 1.81 mL

final burette reading = 39.7 mL

volume of NaOH used = 39.7 - 1.81 = 37.89 mL .  

37.89 mL of .1029 M NaOH is used to neutralise triprotic acid

No of moles contained by 37.89 mL of .1029 M NaOH

= .03789 x .1029 moles

= 3.89 x 10⁻³ moles

Since acid is triprotic ,  its equivalent weight = molecular weight / 3

No of moles of triprotic acid = 3.89 x 10⁻³ / 3

= 1.30   x 10⁻³ moles .

8 0
3 years ago
What is the mass of a neutron?<br><br> 1/2,000 amu<br> 1 amu<br> 2,000 amu<br> 1/200 amu
pashok25 [27]

Vas happenin!!



1 amu is the correct answer


Hope this helps


-Zayn Malik
8 0
3 years ago
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Scientists made the following two observations about emission spectra: each element has a unique emission spectrum. atoms emit e
kondaur [170]
<span>In the Bohr model electrons in atoms can occupy allowed orbits where they do not emit energy. Exchange of energy with the surrounding environment occurs only when an electron "jumps" from an orbit to another. Hope this answers the question. Have a nice day.</span>
3 0
3 years ago
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The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

3 0
3 years ago
A rigid container is filled with chlorine gas. The gas has a pressure of 2.75 bar. The tank is then cooled down to -20.0oC at wh
Gnom [1K]

Answer:

Original temperature (T1) = - 37.16°C

Explanation:

Given:

Gas pressure (P1) = 2.75 bar

Temperature (T2) = - 20°C

Gas pressure (P2) = 1.48 bar

Find:

Original temperature (T1)

Computation:

Using Gay-Lussac's Law

⇒ P1 / T1 = P2 / T2

⇒ 2.75 / T1 = 1.48 / (-20)

⇒ T1 = (2.75)(-20) / 1.48

⇒ T1 = -55 / 1.48

⇒ T1 = - 37.16°C

Original temperature (T1) = - 37.16°C

3 0
3 years ago
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