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worty [1.4K]
3 years ago
7

When you convert a measurement from SI to English, what changes? units , values , error , or resolution

Chemistry
1 answer:
Julli [10]3 years ago
6 0
The answer to your question is units and values
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Do this Q7 if someone do I will her or him brainliest
Simora [160]

The average speed :

1. 10.44 m/s

2. 10.42 m/s

3. 9.26 m/s

The distance 100 m have the greatest average speed

<h3>Further explanation </h3>

Given

Distance and time of runner

Required

Average speed

Solution

<em> Average speed : total distance : total time </em>

1. d = 100 m, t = 9.58 s

Average speed : 100 : 9.58 = 10.44 m/s

2. d=200 m, t=19.19 s

Average speed : 200 : 19.19 = 10.42 m/s

3. d=400 m, t = 43.18 s

Average speed : 400 : 43.18 = 9.26 m/s

The distance 100 m have the greatest average speed

7 0
2 years ago
What happened to the amplitude from wave A to wave B?
dangina [55]
The amplitude doubled
4 0
3 years ago
Read 2 more answers
How many moles of water are in a beaker with 50 mL?
tatiyna

Answer:

Number of moles = 2.8 mol

Explanation:

Given data:

Number of moles of water = ?

Volume of water = 50 mL

Density of water = 1.00 g/cm³

Solution:

1 cm³ =  1 mL

Density = mass/ volume

1.00 g/mL = mass/ 50 mL

Mass = 1.00 g/mL× 50 mL

Mass = 50 g

Number of moles of water:

Number of moles = mass/molar mass

Number of moles = 50 g / 18 g/mol

Number of moles = 2.8 mol

5 0
3 years ago
The value of ka for nitrous acid (hno2) at 25 ∘c is 4.5×10−4. what is the value of δg at 25 ∘c when [h+] = 5.9×10−2m , [no2-] =
LuckyWell [14K]
To get the value of ΔG we need to get first the value of ΔG°:

when ΔG° = - R*T*㏑K

when R is constant in KJ = 0.00831 KJ

T is the temperature in Kelvin = 25+273 = 298 K

and K is the equilibrium constant = 4.5 x 10^-4

so by substitution:

∴ ΔG° = - 0.00831 * 298 K * ㏑4.5 x 10^-4

        = -19 KJ

then, we can now get the value of ΔG when:

ΔG = ΔG° - RT*㏑[HNO2]/[H+][NO2]

when ΔG° = -19 KJ

and R is constant in KJ = 0.00831 

and T is the temperature in Kelvin = 298 K

and [HNO2] = 0.21 m & [H+] = 5.9 x 10^-2 & [NO2-] = 6.3 x 10^-4 m  

so, by substitution:

ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )

       = -40

     
8 0
2 years ago
What are the products of neutralization in Ca(OH)2 + H2CO3
Alinara [238K]

Answer:

CaCO3 + H2O

Explanation:

4 0
3 years ago
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