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Paladinen [302]
3 years ago
9

Quick electron emissions are called

Chemistry
1 answer:
vivado [14]3 years ago
5 0

Answer:

<u>Beta</u><u> </u><u>decay</u>

Quick electron emissions are called <u>beta</u><u> </u><u>decay</u>

<em>Hope</em><em> this</em><em> helps</em>

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3. Fill in: Name the organelle or organelles that perform each of the following functions.
Mekhanik [1.2K]

A. Chloroplasts

B. The cell wall and the vacuole

C. Vacuoles

D. The mitochondrion

8 0
3 years ago
How can you tell if glucose is present?
rosijanka [135]

Answer:

Plants use photosynthesis to make glucose. Glucose is also know as sugar. You can tell it is present if the plant receives sunlight as well as water.

Explanation:

MARK BRAINLIEST

4 0
3 years ago
In one step in the synthesis of the insecticide Sevin, naphthol reacts with phosgene as shown.
Serhud [2]

Answer:

Explanation:

Chemical equation:

C₁₀H₈O + COCl₂  → C₁₁H₇O₂Cl + HCl

A. How many kilograms of C₁₁H₇O₂Cl form from 2.5×10*2 kg of naphthol?

Given data:

Mass of naphthol = 2.5 ×10² kg ( 250×1000 = 250000 g)

Mass of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 250000 g/ 144.17 g/mol

Number of moles of naphthol = 1734.1 mol

Now we will compare the moles of naphthol with C₁₁H₇O₂Cl.

                     C₁₀H₈O       :         C₁₁H₇O₂Cl

                        1               :                1

                       1734.1         :             1734.1

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass = 1734.1 mol × 206.5 g/mol

Mass = 358091.65 g

Gram to kilogram:

1 kg×358091.65 g/ 1000 g  = 358.1 kg

B. If 100. g of naphthol and 100. g of phosgene react, what is the theoretical yield of C11H7O2Cl?

Given data:

Mass of naphthol = 100 g

Mass of COCl₂ = 100 g

Theoretical yield of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 100 g/ 144.17 g/mol

Number of moles of naphthol = 0.694 mol

Number of moles of phosgene:

Number of moles  = mass/ molar mass

Number of moles =  100 g/ 99 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of naphthol and phosgene with C₁₁H₇O₂Cl.

                     C₁₀H₈O        :         C₁₁H₇O₂Cl

                        1                :                1

                       0.694        :              0.694

                    COCl₂          :             C₁₁H₇O₂Cl

                        1                :                1

                       1.0              :              1.0

The number of moles of C₁₁H₇O₂Cl produced by C₁₀H₈O are less so it will limiting reactant and limit the yield of  C₁₁H₇O₂Cl.

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass =  0.694 mol × 206.5 g/mol

Mass = 143.3 g

Theoretical yield  =  143.3 g

C. If the actual yield of C11H7O2Cl in part b is 118 g, what is the percent yield?

Given data:

Actual yield of C₁₁H₇O₂Cl = 118 g

Theoretical yield = 143.3 g

Percent yield = ?

Solution:

Formula :

Percent yield = actual yield / theoretical yield × 100

Now we will put the values in formula.

Percent yield = 118 g/ 143.3 g × 100

Percent yield = 0.82 × 100

Percent yield = 82%

5 0
3 years ago
22.5 g of silver nitrate reacts with excess magnesium bromide, determine the mass
Setler [38]

Answer:

9.82 g of Mg(NO₃)₂

Explanation:

Let's determine the reaction:

2AgNO₃  +  MgBr₂  → Mg(NO₃)₂  +  2AgBr

2 moles of nitrate silver reacts with MgBr₂ in order to produce 1 mol of magnesium nitrate and silver bromide.

We determine the moles of AgNO₃

22.5 g . 1mol / 169.87g = 0.132 moles

Ratio is 2:1.

2 moles of silver nitrate can produce 1 mol of magnesium nitrate

Then, our 0.132 moles may produce (0.132 . 1)/ 2 = 0.0662 moles

We convert moles to mass:

0.0662 mol . 148.3 g/ mol = 9.82 g

6 0
3 years ago
Does a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together? Write the n
olganol [36]

Answer : Yes, a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together.

The net ionic equation will be,

2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

CaCl_2(aq)+2HgNO_3(aq)\rightarrow Ca(NO_3)_2(aq)+Hg_2Cl_2(s)

The ionic equation in separated aqueous solution will be,

Ca^{2+}(aq)+2Cl^{-}(aq)+2Hg^{+}(aq)+2NO_3^{-}(aq)\rightarrow Hg_2Cl_2(s)+Ca^{2+}(aq)+2NO_3^{-}(aq)

In this equation, Ca^{2+}\text{ and }NO_3^- are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)

7 0
3 years ago
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