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ipn [44]
3 years ago
13

What scientist determined that atomic mass did not determine the pattern of the periodic table? A. James Watson B. Henry Moseley

C. Rosalind Franklin D. Dmitri Mendeleev
Chemistry
1 answer:
marta [7]3 years ago
5 0
The answer is B. Henry Moseley
You might be interested in
. ¿Qué cantidad de HNO3 concentrado y de solvente se utilizará en la preparación de 500 mLal 5% (v/v)?
Viefleur [7K]

Answer:

464.29mL de solvente y 35.71mL de ácido nítrico concentrado deben agregarse.

Explanation:

El ácido nítrico concentrado viene al 70% v/v por temas de estabilidad. El volumen de ácido nítrico que se debe agregar si se quieren hace 500mL al 5% de HNO3 es:

500mL * (5mL / 100mL) = 25mL de ácido nítrico se deben agregar.

Como el ácido nítrico está al 70%:

25mL ácido nítrico * (100mL / 70mL ácido nítrico) = 35.71mL de ácido nítrico concentrado deben agregarse.

Y el volumen de solvente debe ser:

500mL - 35.71mL = 464.29mL

6 0
3 years ago
The first excited state of Ca is reached by absorption of 422.7 nm light. Find the energy difference (kJ/mole) between the groun
dedylja [7]

For the excited state of Ca at the absorption of 422.7 nm light,the energy difference  is mathematically given as

E= 4.70x10-22 kJ/mol

<h3>What is the energy difference (kJ/mole) between the ground and the first excited state?</h3>

Generally, the equation for the Energy  is mathematically given as

E = nhc / λ

Where

h= plank's constant

h= 6.625x 10-34 Js

c = speed of light

c= 3x 108 m/s

Therefore

E = 1*(6.625x 10-34 Js)( 3x 10^8 m/s) / ( 422.7x10^-9)

E= 4.70x10-22 kJ/mol

In conclusion, Energy  

E= 4.70x10-22 kJ/mol

Read more about Energy

brainly.com/question/13439286

5 0
2 years ago
You would be most likely to use a slicing machine if you were using the __________ method to produce cookies.
Verizon [17]

You would be most likely to use a slicing machine if you were using the <u>icebox </u>method to produce cookies.

In the icebox method a type of cookie in which the dough is made, rolled into a stick, and refrigerated until the dough hardens. The dough can be removed from the refrigerator, cut into individual pieces, and then baked. The rest of the dough is returned to the refrigerator until needed.

Icebox method, also known as refrigerator cookies, are sliced ​​and baked cookies. The dough is formed into logs, chilled in the refrigerator (also called an icebox), sliced ​​, and then baked.

Learn more about the Icebox method here,

https://brainly.in/question/1513677

#SPJ4

7 0
1 year ago
The balanced equation below shows the products that are formed when butane (C4H10) is combusted.
Nataly [62]

Answer:

2:8

Explanation:

The reaction equation is a given as:

         2C₄H₁₀   +    130₂   →    8CO₂     +     10H₂O  

From the reaction equation, the mole ratio is 2:8

Butane is C₄H₁₀

Carbon dioxide CO₂

From the reaction;

       2 moles of butane will produce 8 moles of carbon dioxide

3 0
3 years ago
Read 2 more answers
During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of
almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate \rm CaCO_3?

Refer to a modern periodic table for relative atomic mass data:

  • Ca: 40.078;
  • C: 12.011;
  • O: 15.999.

Formula mass of \rm CaCO_3:

M(\mathrm{CaCO_3})  = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}.

\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol.

How many moles of \rm CaCl_2 will be produced?

The coefficient in front of \rm CaCO_3 in the chemical equation is the same as that in front of \rm CaCl_2. That is:

\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1.

\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol.

What's the theoretical yield of calcium chloride? In other words, what's the mass of \rm 0.949184\;mol of \rm CaCl_2?

Again, refer to a periodic table for relative atomic data:

  • Ca: 40.078;
  • Cl: 35.45.

M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}.

\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}.

What's the actual yield of calcium chloride?

\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%.

\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}.

8 0
3 years ago
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