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kicyunya [14]
3 years ago
11

Which animal can leave its eggs to hatch and does not need to take care of the infants?

Chemistry
2 answers:
Komok [63]3 years ago
7 0
Sea turtles, penguins hold them in a pouting, ducks keep them around sea turtles leave them
Murljashka [212]3 years ago
4 0
Sea turtles they leave them on the beach
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how many moles of ammonium chloride are required for a complete reaction with 420 mol of calcium hydroxide
Oxana [17]

3 KOH + H3PO3 = K3PO4 + 3 H2O  

moles KOH = 26.5 g/56.1 g/mol=0.472  

moles K3PO4 = 0.472/3 =0.157

3 0
3 years ago
At a certain temperature the rate of this reaction is first order in SO₃ with a rate constant of 0.208 s⁻¹:
NNADVOKAT [17]

Answer:

0.30 M

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t  = ?

[A_0] is the initial concentration  = 1.36 M

k is the rate constant = 0.208 s⁻¹

t = 7.30 seconds

So,  

[A_t]=1.36\times e^{-0.208\times 7.30}\ M=0.30\ M

8 0
3 years ago
What is the relative rate of diffusion of hydrogen and nitrogen ​
vredina [299]

Answer:

2.645

Explanation:

Rate of diffusion formula:

Sqrt(mass2/mass1)

>>sqrt(14/2)

(Note:Hydrogen must exist in dwiatomic, [H2])

3 0
2 years ago
The piece of iron that miguel measured had a mass of 51.1 g and a volume of 6.63 cm 3 . what did miguel calculate to be the dens
likoan [24]
Given:
Mass, m = 51.1 g
Volume, V = 6.63 cm³

By definition, 
Density = Mass/Volume
              = (51.1 g)/(6.63 cm³)
              = 7.7074 g/cm³

In SI units,
Density = (7.7074 g/cm³)*(10⁻³ kg/g)*(10² cm/m)³
              = 7707.4 kg/m³

Answer: 7.707 g/cm³ or 7707.4 kg/m³

4 0
3 years ago
Boron has two naturally-ocurring isotopes. Boron-10 has an abundance of 19.8% and actual mass of 10.013 amu, and boron-11 has an
aniked [119]

Answer:

Average atomic mass = 10.812 amu

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

<u>For first isotope, Boron-10: </u>

% = 19.8 %

Mass = 10.013 amu

<u>For second isotope, Boron-11: </u>

% = 80.2 %

Mass = 11.009 amu

Thus,  

Average\ atomic\ mass=\frac{19.8}{100}\times {10.013}+\frac{80.2}{100}\times {11.009}=1.982574+8.829218=10.811792

<u>Average atomic mass = 10.812 amu</u>

6 0
3 years ago
Read 2 more answers
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