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WINSTONCH [101]
3 years ago
13

Help ASAP!!!!!!!!!!!!!!!!!!!!!!!!!

Chemistry
1 answer:
BartSMP [9]3 years ago
3 0

Answer:

ok      .

Explanation:            

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What pattern appears in the vertical arrangement (columns) of the elements in the Periodic Table, moving from top to bottom?
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Idk if this is the information you need but the energy level gets higher the farther you move down the Periodic Table. Every element in a vertical line has the same energy level.
3 0
3 years ago
The temperature of a gas is 30 degrees Celsius and its pressure is 760 torr. If the temperature originally
erastova [34]
T₁ = 40°C + 273.15 = 313.15 Kelvin T₂ = 30°C + 273.15 = 303.15 Kelvin

Solving Gay-Lussac's Law for P₁ we get:
P₁ = P₂ • T₁ ÷ T₂ P₁ = 760 torr • 313.15 K ÷ 303.15 K P₁ = 785.07 torr

Using the calculator, we click on the P1 button.
We then enter the 3 numbers 760 313.15 and 303.15 into the correct boxes then click "CALCULATE" and get our answer of 785.07 torr.
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How is DNA related to chromosomes????????????:/
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3 years ago
Read 2 more answers
CalcuLate the velocity of an electron ejected if 300.0 mm of light is applied to the surface. A wavelength of 795 nm has suffici
worty [1.4K]
The equation relating velocity and wavelength is written below:

v = λf
where λ is the wavelength in m while f is frequency in 1/s.

Let's determine first the frequency from the speed of light:
c = distance/time, where c is the speed of light equal to 3×10⁸ m/s
3×10⁸ m/s = (300 mm)(1 m/1000 mm)/ time
time = 1×10⁻⁹ seconds
Since f = 1/t,
f = 1/1×10⁻⁹ seconds = 10⁹ s⁻¹

Thus,
v = (795×10⁻⁹ m)(10⁹ s⁻¹)
v = 795 m/s
8 0
3 years ago
You are given a sulfuric acid solution of unknown concentration. You dispense 10.00 mL of the unknown solution into an Erlenmeye
dmitriy555 [2]

Answer:

"0.053457 M" of sulfuric acid.

Explanation:

The given values are:

V = 10 mL solution

V_{added} = 12.20 mL

V_{total} = 22.20 mL

then,

M 0.103 M of NaOH,

V_{rinsed} = experiment  will not be affected

V_{total \ base} = 10.38 mL

Now,

⇒  mol of NAOH = MV

                            = 0.103\times 10.38

                            =  1.06914  \ m

Whether Sulfuric acid, then

⇒  H_{2}SO_{4} + 2NaOH = Na_{2}SO_{4} + 2H_{2}O

⇒  mol \ of \ acid =\frac{1}{2}\times \ mol \   of  \ base

⇒  1.06914 \ m \ mol \ of \ base = \frac{1}{2}\times 1.06914 = 0.53457 \ m \ mol \ of \ acid

Before any dilution:

V_{sample} = 10  \ mL

⇒  M \ acid = \frac{m \ mol}{V}

                 =\frac{ 0.53457 }{10}

                 =0.053457 \ M (Sulfuric acid)

6 0
3 years ago
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