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3 years ago

I need helppp(science)

Chemistry

1 answer:

Mama L [17]3 years ago

7 0

Answer:

Flat plains because when the magma flies out the spews out the volcano it will sinks to the ocean floor it rise and then once more eruptions come the magma will form with land that was previously there.

Explanation:

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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

What is Minimum Threshold Frequency?

hoa [83]

the minimum frequency of radiation that will produce a photoelectric effect

<h2>#CarryOnLearning</h2>

7 0

3 years ago

Read 2 more answers

When iron metal reacts with oxygen, the reaction can form Fe2O3. Write a balanced chemical equation for this reaction, and find

I am Lyosha [343]

<h3>Answer:</h3>

Balanced equation: 4Fe + 3O₂ → 2Fe₂O₃

Moles of oxygen gas = 9 moles

<h3>Explanation:</h3>

To answer the question;

We first write the balanced equation between iron metal and Oxygen

The balanced equation is given as;

4Fe + 3O₂ → 2Fe₂O₃

We are given 6 moles of Fe₂O₃

We are required to determine the number of moles of oxygen needed to form 6 moles of Fe₂O₃.

From the equation, 3 moles of oxygen gas reacts to produce 2 moles of Fe₂O₃

This means, the mole ratio of O₂ to Fe₂O₃ is 3 : 2

Therefore; Moles of O₂ = Moles of Fe₂O₃ × 3/2

Hence, moles of oxygen = 6 moles × 3/2

= 9 moles

Thus, Moles of Oxygen needed is 9 moles

6 0

3 years ago

If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W

olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

$CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2$

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

$m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl$

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

$m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3$

Therefore, the leftover of sodium nitrate is:

$m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3$

Finally, the percent yield is computed via:

$Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%$

Best regards!

6 0

3 years ago

Using the balanced equation for the next few questions: 4 Fe(s) + 30 (9) - 2Fe2O3(s)

stira [4]

Answer:

0.075 moles of iron oxide would be produced by complete reaction of 0.15 moles of iron.

Explanation:

The balanced reaction is:

4 Fe + 3 O₂ → 2 Fe₂O₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Fe: 4 moles
O₂: 3 moles
Fe₂O₃: 2 moles

You can apply the following rule of three: if by stoichiometry 4 moles of Fe produce 2 moles of Fe₂O₃, 0.15 moles of Fe produce how many moles of Fe₂O₃?

$moles of Fe_{2} O_{3} =\frac{0.15 moles of Fe*2 moles of Fe_{2} O_{3} }{4 moles of Fe}$

moles of Fe₂O₃= 0.075

<u><em>0.075 moles of iron oxide would be produced by complete reaction of 0.15 moles of iron.</em></u>

5 0

3 years ago