All categories

3 years ago

Does salt form a homogeneous mixture when mixed with water

2 answers:

6 0

Yes because a homogeneous mixture is a mixture that has the same uniform appearance and composition throughout while with heterogeneous you can see the different parts of the mixture and pick them out

igomit [66]3 years ago

3 0

Answer:

Yes

Explanation:

I got this right on the test

You might be interested in

What would be the most likely outcome if the ribosomes of a cell were damaged and stopped functioning?

finlep [7]

Greetings!

Option A would be correct. The most likely outcome, (if) ribosomes stopped functioning properly, would be Cell growth stopping, or slowing down.

~Lauv

5 0

3 years ago

Read 2 more answers

The anser is A trust me.....

natulia [17]

Answer:

Explanation:

You said 'The anser is A trust me.....'

8 0

2 years ago

Are involuntary muscles striated?

alukav5142 [94]

No.The involuntary muscles of the muscular system are not striated or furrowed.


These muscles that are not striated are very smooth to the sight which is why they are also called Smooth muscles. Involuntary muscles are not striated. Even among the striated muscles that are voluntary, the heart which is also a muscle, is involuntary.
On the other hand the striated muscles, which are voluntary are highly fibrous and their primary function is to move the body voluntarily, and producing heat as a result of this muscle action.

4 0

3 years ago

How many chromosomes does a human cell contain at the start of mitosis? How many chromosomes does a human cell contain at the en

Umnica [9.8K]

Answer:

46 chromosomes

two chromosomes created at the end of cytokinesis

Explanation:

6 0

3 years ago

In flies dumpy wings and ebony bodies are each mutations recessive to wild type, and you expect them to behave in a Mendelian fa

Dimas [21]

Answer:

See the answers below

Explanation:

Assuming that D (d) represents the allele for wing type and E (e) represents the allele for body type, crossing DdEe with DdEe will yield the following offspring according to the Punnet's square (see the attached image):

(a) 9/16 D_E_ wild type

3/16 D_ee wild wing, ebony body

3/16 ddE_ dumpy wing, wild body

1/16 ddee dumpy wing, ebony body

(b) Chi square $X^2$ = $\frac{(O - E)^2}{E}$ , where O = observed frequency and E = expected frequency.

Phenotype O E $X^2$

Wild type 473 9/16 x 830 = 466.875 $\frac{(473 - 466.875)^2}{466.875}$ = 0.08

wild w/ebony b 156 3/16 x 830 = 155.625 $\frac{(156 - 155.625)^2}{155.625}$ = 0.0009

Dumpy w/wild b 149 3/16 x 830 = 155.625 $\frac{(149 - 155.625)^2}{155.625}$ = 0.28

dumpy w/ebony b 52 1/16 x 830 = 51.875 $\frac{(52 - 51.875)^2}{51.875}$ = 0.0003

Total $X^2$ = 0.3612

Degree of freedom = 4 - 1 = 3

Tabulated value of $X^2$ (0.05)= 7.815

(c) Since the calculated Chi square value is less than the tabulated value, we conclude that the observed outcome agrees with the expected outcome and that the cross followed the standard Mendelian pattern.

6 0

3 years ago