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3 years ago

5

In Rutherford's golf foil experiment, no a particles stuck to the atom. If a particles had stuck to the atom, what might you hav

e
deduced about the atom's structure?

1 answer:

borishaifa [10]3 years ago

4 0

Answer

:yes or no

Explanation:

you needd tol try it yolurself

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help!! i tried answering these on my own and turning it into my teacher and she said they were all wrong , can somebody help me

Kobotan [32]

I don't know what your answers were but the more "stuff" there is the more energy it takes to freeze or boil it. Hope this could help

6 0

3 years ago

How much energy is required to heat 40g of water from -7 degrees Celsius to 108 degrees Celsius (Lf= 335,000J/Kg for ice)

solniwko [45]

The energy required to heat 40g of water from -7 c to 108 c is
1541000 joules

calculation

Q(heat)= M( mass) x c(specific heat capacity) xdelta t( change in temperature)

M= 40g= 40/1000= 0.04 Kg
C= 335,000 j/kg/c
delta T ( 108 --7= 115 c)

Q is therefore = 0.04 g x 335000 j/kg/c x 115 c = 1541,000 joules

6 0

3 years ago

6) (a) Calculate the absorbance of the solution if its concentration is 0.0278 M and its molar extinction coefficient is 35.9 L/

Anvisha [2.4K]

Answer:

6) (a) 0.499; (b) 31.7 %

7) 0.15

Explanation:

6) (a) Absorbance

Beer's Law is

$A = \epsilon cl\\A = \text{35.9 L&\cdot$mol$^{-1}$cm$^{-1}$} $\times$ 0.0278 mol$\cdot$L$^{-1} \times $ 0.5 cm = \mathbf{0.499}$

(b) Percent transmission

$A = \log {\left (\dfrac{1}{T}}\right)}\\\\\%T = 100T\\\\T = \dfrac{\%T}{100}\\\\\dfrac{1}{T} = \dfrac{100 }{\%T}\\\\A = \log \left(\dfrac{100 }{\%T} \right ) = 2 - \log \%T\\\\0.499 = 2 - \log \%T\\\\\log \%T = 2 - 0.499 = 1.501\\\\\%T = 10^{1.501} = \mathbf{31.7}$

7) Absorbance

$A = \log \left (\dfrac{I_{0}}{I} \right ) = \log \left (\dfrac{I_{0}}{0.70I_{0}} \right ) = \log \left (\dfrac{1}{0.70} \right ) = -\log(0.70) = \mathbf{0.15}}$

8 0

3 years ago

How many grams of H2O are produced when 35.0 g of NaOH reacts with 17.5 g of CO,?

zhenek [66]

Answer:

2NaOH + CO2 -> Na2CO3 + H2O

1) Find the moles of each substance

$\eq n(NaOH)=\frac{35.0}{22.99+16.00+1.008\\}\ =\frac{35.0}{39.998} \ = 0.8750437522 moles\\n(CO_{2} ) = \frac{17.5}{12.01+32.00} = \frac{17.5}{44.01} = 0.3976369007 moles\\$

2) Determine the limitting reagent

$\\NaOH = \frac{0.8750437522}{2} = 0.4375218761\\\\$

∴ Carbon dioxide is limitting as it has a smaller value.

3) multiply the limiting reagent by the mole ratio of unknown over known

n(H2O ) = 0.3976369007 × 1/2

= 0.1988184504 moles

4) Multiply the number of moles by the molar mass of the substance.

m = 0.1988184504 × (1.008 × 2 + 16.00)

= 0.1988184504 × 18.016

= 3.581913202 g

Explanation:

6 0

2 years ago

Which of the following is NOT one of the three main global climate zones?

krok68 [10]

The answer is Latitude (B)

8 0

2 years ago