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lina2011 [118]
3 years ago
12

Please select the word from the list that best fits the definition growth from seed to plant

Chemistry
2 answers:
Misha Larkins [42]3 years ago
8 0
Is there a picture for this question ?
NeX [460]3 years ago
5 0

Answer:

kay, so u didn't gave the list but I guess I'm right...bcz of some words you wrote below..

yeah,

so your answer is <u>GERMINATION</u>

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In which type of chemical reaction do two or more reactants combine to form one product, only?
xenn [34]
In decomposition, two or more reactant combine to form one product only.
8 0
3 years ago
How many grams of Na2SO4 should be weighed out to prepare 0.5L of a 0.100M solution?​
Nezavi [6.7K]

Answer:

7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100 M solution.

Explanation:

First of all the molecular weight of Na2SO4 is 142.08 gram.Now we all know that if the molecular weight of a compound is dissolved in 1000ml or 1 litee of water then the strength of that solution becomes 1 M.

    According to the given question we have to prepare 0.100 M solution

1000 ml of solution contain 142.08×0.1= 14.208 gram Na2SO4

1    ml of solution contain     14.208÷1000= 0.014 gram

0.5L or 500ml of solution contain 0.014×500= 7gram Na2SO4.

 So it can be stated that 7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100M solution.

     

3 0
3 years ago
What is the net ionic equation for the reaction between nickle (II) chloride and silver (I) nitrate?
weqwewe [10]

Answer:

Down below

Explanation:

The following uses nickel(II) chloride

2AgNO3(aq) + NiCl2(aq) ==> Ni(NO3)2(aq) + 2AgCl(s) Molecular

8 0
3 years ago
It takes 167 s for an unknown gas to effuse through a porous wall and 99 s for the same volume of n2 gas to effuse at the same t
irakobra [83]

Answer : The molar mass of the unknown gas will be 79.7 g/mol


Explanation : To solve this question we can use graham's law;


Now we can use nitrogen as the gas number 2, which travels faster than gas 1;

So, 167 / 99 = 1.687 So the nitrogen gas is 1.687 times faster that the unknown gas 1

We can compare the rates of both the gases;


So here, Rate of gas 2 / Rate of gas 1 = \sqrt{(molar mass 1 / molar mass 2)}

Now, 1.687 = square root [\sqrt{(molar mass 1) / (28.01 g/mol N_{2})} ]


When we square both the sides we get;


2.845 = (molar mass 1) / (28.01 g/mol N2)


On rearranging, we get,


2.845 X (28.01 g/mol N2) = Molar mass 1

So the molar mass of unknown gas will be = 79.7 g/mol

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
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