How does active transport differ from passive transport

The trait for flower color in a plant has red and white alleles. The red color is the dominant trait. What is the phenotypic rat

Fudgin [204]

Answer :

2 red : 2 white

Explanation:

;)long story short

hetro with hetro gives 3:1
hetro with recessive gives 1:1

Hetro :- ( Rr) one capital letter and one small
Recessive :- (rr) two small leters

6 0

2 years ago

Draw the lewis structure for butanal, which has the condensed formula ch3(ch2)2cho. show all hydrogen atoms and lone pairs of el

sergejj [24]

There are 10 hydrogen atoms that bind and there are 2 pairs of free electrons in the non-binding O atom

<h3>Further explanation</h3>

Aldehydes are alkane-derived compounds containing carbonyl groups (-CO-) where one bond binds to an alkyl group while another binds to a hydrogen atom.

The general structure is R-CHO with the molecular formula :

$\large{\boxed{\boxed{\bold{C_nH_{2n}O}}}$

Naming is generally the same as the alkane by replacing the suffix with -al

Butanal or butyraldehyde is an aldehyde which has 4 C atoms

Inside the structure there are 3 atoms involved in bonding:

1. Atom C with 4 valence electrons, requires 4 electrons to reach the octet

2. Atom O with 6 valence electrons, requires 2 electrons to reach the octet

3. Atom H with 1 valence electron, requires 1 electron to reach a duplet

In describing Lewis's structure the steps that can be taken are:

1. Count the number of valence electrons from atoms in a molecule

2. Give each bond a pair of electrons

3. The remaining electrons are given to the atomic terminal so that an octet is reached

4. The remaining electrons that still exist in the central atom

5. If the central atom is not yet octet, free electrons are drawn to the central atom to form double bonds

In the Butanal structure (C₄H₈O) there is 1 double bond of the functional group (-CHO) between the C atom and the O atom

<h3>Learn more: </h3>

Adding electron dots

brainly.com/question/6085185

Ionic bonding

brainly.com/question/1603987

Formal charge

brainly.com/question/7190235

Keywords: butanal, aldehyde, Lewis structure, a valence electron

8 0

3 years ago

Read 2 more answers

When the reaction shown is correctly balanced, the coefficients are: kclo3 → kcl + o2?

notka56 [123]

From the balanced equation 2KClO3 → 2KCl + 3O2, the coefficients are the following:
coefficient 2 in front of potassium chlorate KClO3
coefficient 2 in front of potassium chloride KCl
coefficient 3 in front of oxygen molecule O2

We got this balanced equation by identifying the number of atoms of each element that we have in the given equation KClO3 → KCl + O2.
Looking at the subscripts of each atom on the reactant side and on the product side, we have
KClO3 → KCl + O2
   K=1 K=1
Cl=1 Cl=1
O=3 O=2

We can see that the oxygens are not balanced. We add a coefficient 2 to the 3 oxygen atoms on the left side and another coefficient 3 to the 2 oxygen
atoms on the right side to balance the oxygens:
2KClO3 → KCl + 3O2
The coefficient 2 in front of potassium chlorate KClO3 multiplied by the subscript 3 of the oxygen atoms on the left side indicates 6 oxygen atoms just as the coefficient 3 multiplied by the subscript 2 on the right side indicates 6 oxygen atoms.

The number of potassium K atoms and chloride Cl atoms have changed as well:
2KClO3 → KCl + 3O2
   K=2 K=1
   Cl=2 Cl=1
    O=6 O=6

We now have two potassium K atoms and two chloride Cl atoms on the reactant side, so we add a coefficient 2 to the potassium chloride KCl on the product side:
2KClO3 → 2KCl + 3O2, which is our final balanced equation.
K=2 K=2
Cl=2 Cl=2
O=6 O=6
The potassium, chlorine, and oxygen atoms are now balanced.

5 0

2 years ago

Read 2 more answers

A sample of 10.5 g of nitrogen reacts with 20.2 g of hydrogen to produce ammonia. (36 points

BartSMP [9]

Answer:

Explanation:

A. 55

B. 43

C. 14

D. 29

6 0

1 year ago

What is the molar out of a solution that contains 33.5g of CaCl2 in 600.0mL of water

omeli [17]

Answer:

Here's what I got.

Explanation:

Interestingly enough, I'm not getting

0.0341% w/v

either. Here's why.

Start by calculating the percent composition of chlorine,

Cl

, in calcium chloride, This will help you calculate the mass of chloride anions,

Cl

−

, present in your sample.

To do that, use the molar mass of calcium chloride, the molar mass of elemental chlorine, and the fact that

1

mole of calcium chloride contains

2

moles of chlorine atoms.

2

×

35.453

g mol

−

1

110.98

g mol

−

1

⋅

100

%

=

63.89% Cl

This means that for every

100 g

of calcium chloride, you get

63.89 g

of chlorine.

As you know, the mass of an ion is approximately equal to the mass of the neutral atom, so you can say that for every

100 g

of calcium chloride, you get

63.89 g

of chloride anions,

Cl

−

.

This implies that your sample contains

0.543

g CaCl

2

⋅

63.89 g Cl

−

100

g CaCl

2

=

0.3469 g Cl

−

Now, in order to find the mass by volume percent concentration of chloride anions in the resulting solution, you must determine the mass of chloride anions present in

100 mL

of this solution.

Since you know that

500 mL

of solution contain

0.3469 g

of chloride anions, you can say that

100 mL

of solution will contain

100

mL solution

⋅

0.3469 g Cl

−

500

mL solution

=

0.06938 g Cl

−

Therefore, you can say that the mass by volume percent concentration of chloride anions will be

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the volume of the solution.

.

ALTERNATIVE APPROACH

Alternatively, you can start by calculating the number of moles of calcium chloride present in your sample

0.543

g

⋅

1 mole CaCl

2

110.98

g

=

0.004893 moles CaCl

2

To find the molarity of this solution, calculate the number of moles of calcium chloride present in

1 L

=

10

3

mL

of solution by using the fact that you have

0.004893

moles present in

500 mL

of solution.

10

3

mL solution

⋅

0.004893 moles CaCl

2

500

mL solution

=

0.009786 moles CaCl

2

You can thus say your solution has

[

CaCl

2

]

=

0.009786 mol L

−

1

Since every mole of calcium chloride delivers

2

moles of chloride anions to the solution, you can say that you have

[

Cl

−

]

=

2

⋅

0.009786 mol L

−

1

[

Cl

−

]

=

0.01957 mol L

−

This implies that

100 mL

of this solution will contain

100

mL solution

⋅

0.01957 moles Cl

−

10

3

mL solution

=

0.001957 moles Cl

−

Finally, to convert this to grams, use the molar mass of elemental chlorine

0.001957

moles Cl

−

⋅

35.453 g

1

mole Cl

−

=

0.06938 g Cl

−

Once again, you have

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

In reference to the explanation you provided, you have

0.341 g L

−

1

=

0.0341 g/100 mL

=

0.0341% m/v

because you have

1 L

=

10

3

mL

.

However, this solution does not contain

0.341 g

of chloride anions in

1 L

. Using

[

Cl

−

]

=

0.01957 mol L

−

1

you have

n

=

c

⋅

V

so

n

=

0.01957 mol

⋅

10

−

3

mL

−

1

⋅

500

mL

n

=

0.009785 moles

This is how many moles of chloride anions you have in

500 mL

of solution. Consequently,

100 mL

of solution will contain

100

mL solution

⋅

0.009785 moles Cl

−

500

mL solution

=

0.001957 moles Cl

−

So once again, you have

0.06938 g

of chloride anions in

100 mL

of solution, the equivalent of

0.069% m/v

.

Explanation:

i think this is it

8 0

2 years ago