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3 years ago

1. Which type of metal did Rutherford shoot alpha particles into?

1 answer:

7 0

<span>1. Which type of metal did Rutherford shoot alpha particles into?
C)Gold

2.When Thomson held a positively charged plate near the cathode ray, the beam bent toward the plate. What conclusion can be drawn from this observation?
C)The beam was negatively charged.

3.)What was observed during Rutherford's experiment?
D)Most of the particles went through the foil, but some were deflected.</span>

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Question 11 please HELP

Lena [83]

The answer is A or B, I would put B.

6 0

3 years ago

Compared with the freezing-point depression of a 0.01 m c6h12o6 solution, the freezing-point depression of a 0.01 m nacl solutio

agasfer [191]

Answer:

Twice as much.

Explanation:

That's because the freezing point depression depends on the total number of solute particles.

C₆H₁₂O₆(s) ⟶ C₆H₁₂O₆(aq)

0.01 mol of C₆H₁₂O₆ gives 0.01 mol of solute particles.

NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)

1 mol of NaCl gives 0.01 mol of Na⁺(aq) and 0.01 mol of Cl⁻(aq).

That's 0.02 mol of particles, so the freezing point depression of 0.01 mol·L⁻¹ NaCl will be twice that of 0.01 mol·L⁻¹ C₆H₁₂O₆.

6 0

3 years ago

A chemist wants to make 6.5 L of a 0.340 M CaCl2 solution. Part A What mass of CaCl2 (in g) should the chemist use? Express your

katovenus [111]

Answer:

The answer to your question is 245 grams

Explanation:

Data

Volume 6.5 L

Molarity = 0.34

mass of CaCl₂ = ?

Process

1.- Calculate the molar mass of CaCl₂

molar mass = (1 x 40) + (2 x 35.5)

= 40 + 71

= 111 g

2.- Convert the grams to moles

111 g of CaCl₂ -------------- 1 mol

x ---------------0.34 mol

x = (0.34 x 111) / 1

x = 37.74 g

3.- Calculate the total mass

37.74 g ------------------ 1 L

x ------------------ 6.5 L

x = (6.5 x 37.74) / 1

x = 245.31

3 0

3 years ago

Calculate the volume of the acid solution and the volume of the conjugate base solution that would be needed to prepare a buffer

bogdanovich [222]

Answer:

Explanation:

This can be contradictory, depending on whether the 0.1 M

is the total species concentration or the concentration of each of the two components. I'll consider this to be the former...

VA− = 9.125 mL

VHA = 15.875 mL

The Henderson-Hasselbalch equation is:

pH = pKa + log [A−][HA]

We have a pH 4.5

solution of acetic acid and acetate, so from there we can get the ratio of weak acid to conjugate base:

[A−][HA]=10

pH − pKa = 104.5 − 4.74 = 0.5754

Now, if the total concentration is

0.10 M , then:

[HA] + [A−] 0.5754

[HA] = 0.10 M

⇒[HA] = 0.10 M 1.0000 +0.5754

= 0.0635 M

−−−−−−−−

⇒[A−] = 0.0365 M

−−−−−−−−

and these concentrations are AFTER mixing. Since the total volume is 50 mL , or 0.050 L, the mols of each component (which are constant!) are:

nA − = 0.0365 molL × 0.050L =

0.001825 mols

−−−−−−−−−−−−

nHA = 0.0635 molL × 0.050L =

0.003175 mols

−−−−−−−−−−−−

So, if both of the starting concentrations were

0.20 M, we can find the volume they each start with:

VA − = 1 L0.20mols

A− × 0.001825mols A− = 0.009125 L = 9.125 mL

−−−−−−−−

VHA = 1 L 0.20 mols HA × 0.003175

mols HA = 0.015875 L = 15.875 mL

−−−−−−−−−

And this should make sense, because the total starting volume is

25.000 mL , the total ending volume is twice as large; the total species concentration is half the concentration that both species started with.

6 0

3 years ago

If the absorbance of a KMnO4 solution of unknown concentration is 0.633, calculate the concentration of KMnO4 in the solution.

solong [7]

The molar Concentration of KMnO₄ is 0.000219 M

Concentration is the abundance of a constituent divided by means of the overall extent of an aggregate. numerous styles of mathematical description may be outstanding: mass awareness, molar awareness, variety concentration, and quantity awareness.

y is absorbance

x is the molar concentration of KMnO_4

y = 4.84E + 03x - 2.26E - 01

0.833 = 4.84 * 10⁺⁰³ x - 2.26 * 10⁻¹

1.059 = 4.84 * 10⁺⁰³ x

X = 0.000219 M

Hence, The molar Concentration of KMnO₄ is 0.000219 M

Learn more about concentration here:-brainly.com/question/14469428

#SPJ9

3 0

1 year ago