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2 years ago

Number 5 please this is due in 3 hours I’m desperate :((((

Chemistry

2 answers:

Zielflug [23.3K]2 years ago

5 0

The developing zygote is protected would be the correct answer
not 100% sure

ss7ja [257]2 years ago

3 0

#4 )

The option that is a disadvantage of external fertilization is the option that states gametes may get washed away.

#5 )

The option that is an advantage of internal fertilization is the option that states the developing zygote is protected.

You might be interested in

The symbols from the list and match them with the names of the

stich3 [128]

Be - Beryllium
S - sulfur
K - Potassium
C - Carbon
B - Boron
Ar - Argon
O - oxygen
Ne - Neon
Ca - Calcium
H - Hydrogen

3 0

3 years ago

Chlorine has two naturally stable isotopes: 35Cl (34.968853 amu) and 37Cl (36.965903 amu). The natural abundance of each isotope

allochka39001 [22]

Answer:

35Cl ⇒ 34.968853 amu has an abundance of 30.05%

Explanation:

The molar mass of chlorine, (which is the average of all its naturally stable isotope masses), is 36.36575 amu.

There are 2 naturally stable isotopes, this means together they have an abundance of 100%

The isotopes are:

35Cl ⇒ 34.968853 amu has an abundance of X %

37Cl ⇒ 36.965903 amu has an abundance of Y %

X + Y = 100% OR X = 100% - Y

36.36575 = 34.968853X + 36.965903Y

36.36575 = 34.968853(1-Y) + 36.936.96590365903Y

36.36575 = 34.968853 -34.968853Y + 36.965903Y

1.396897 = 1.99705Y

Y = 0.699 = 69.95%

X = 100-69.9 = 30.05%

To control, we can plug in the following equation:

34.968853 * 0.3005 + 36.965903 * 0.6995 = 36.3658

This means

37Cl ⇒ 36.965903 amu has an abundance of 69.95 %

35Cl ⇒ 34.968853 amu has an abundance of 30.05%

7 0

3 years ago

I have 2 samples of solid chalk (aka calcium carbonate). Sample A has a total mass of 4.12 g and Sample B has a total mass of 19

IRINA_888 [86]

Answer:

A) Sample B has more calcium carbonate molecules

Explanation:

M = Molar mass of calcium carbonate = 100.0869 g/mol

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

For the 4.12 g sample

Moles of a substance is given by

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{4.12}{100.0869}\\\Rightarrow n=0.0411\ \text{mol}$

Number of molecules is given by

$nN_A=0.0411\times 6.022\times 10^{23}=2.48\times 10^{22}\ \text{molecules}$

For the 19.37 g sample

$n=\dfrac{19.37}{100.0869}\\\Rightarrow n=0.193\ \text{mol}$

Number of molecules is given by

$nN_A=0.193\times 6.022\times 10^{23}=1.16\times 10^{23}\ \text{molecules}$

$1.16\times 10^{23}\ \text{molecules}>2.48\times 10^{22}\ \text{molecules}$

So, sample B has more calcium carbonate molecules.

The ratio of the elements of carbon, oxygen, calcium atoms, ions, has to be same in both the samples otherwise the samples cannot be considered as calcium carbonate. Same is applicable for impurities. If there are impurites then the sample cannot be considered as calcium carbonate.

7 0

3 years ago

A 0.73 m solution of chloroacetic acid, , has a ph of 1.50. calculate for the acid.

erik [133]

PH = 0.1289 for 1.50 M solution of weak acid with Ka value of .73

6 0

3 years ago

PLEASE HELP!

11111nata11111 [884]

The answer is A.......

6 0

3 years ago