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Goshia [24]
2 years ago
7

Help please I’ll give you brainless

Chemistry
1 answer:
Nikitich [7]2 years ago
8 0

Answer:

By atomic number?

Explanation:

fingers crossed its right :/

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Select the most likely product for this reaction:
Advocard [28]

Answer:

The answer is B- LiNO3(aq)+H2O(I)

Explanation:

I just did the assignment also UNUS ANNUS

3 0
3 years ago
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The diagram shows a model of the nitrogen cycle. which role do plants play in the nitrogen cycle?
kenny6666 [7]

Answer

D

Explanation:

They take up usable forms of nitrogen found in soil

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3 years ago
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naturally occurring bromine molecules, br2 have masses of 158, 160, and 162. they occur in the relative abundances 25.69%, 49.99
4vir4ik [10]
The average atomic mass of an element can be determined by multiplying the individual masses of the isotopes with their respective relative abundances, and adding them. 

Average atomic mass of Br = 158 amu(0.2569) + 160 amu(0.4999) + 162 amu(0.2431)
Average atomic mass = 159.96 amu

As described in the problem, the relative abundance for Br-79 is 25.69%. This is because 2 atoms of Br is equal to 79*2 = 158 amu. Similarly, the relative abundance of Br-81 is 81*2 = 162, which is 24.31%.
4 0
3 years ago
A sample of metal has a mass of 16.5916.59 g, and a volume of 5.815.81 mL. What is the density of this metal
Oxana [17]

Answer:

=> 2.8554 g/mL

Explanation:

To determine the formula to use in solving such a problem, you have to consider what you have been given.

We have;

mass (m)     = 16.59 g

Volume (v) =  5.81 mL

From our question, we are to determine the density (rho) of the rock.

The formula:

p = \frac{m}{v}

Substitute the values into the formula:

​p = \frac{16.59 g}{5.81 mL} \\   = 2.8554 g/mL

= 2.8554 g/mL

Therefore, the density (rho) of the rock is 2.8554 g/mL.

5 0
1 year ago
Calculate the dissociation constant of nh4oh(aq) if the degree of dissociation of 0.006 mol/kg solution is 0.053 and the activit
Anastasy [175]

The dissociation equation will be

                         NH4OH   --->        NH4+                   + OH-

Initial                 0.006                        0                          0

Change         -0.006 X 0.053        +0.006 X 0.053      -0.006 X 0.053

Equlibrium     0.006 -0.006 X 0.053      0.006 X 0.053    0.006 X 0.053

Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053

Ka = 1.78 X 10^-5

7 0
3 years ago
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