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3 years ago

I’ll mark you brainley if you help me figure out which answer it is!!!

Chemistry

1 answer:

Stolb23 [73]3 years ago

5 0

Answer:

Explanation:

A chemical change happens when one chemical substance is transformed into one or more different things , such as when iron becomes rust so

B: Acids turn the indicator red, pink, orange, and yellow, while bases turn it green based on what it reactions with or comes in contact with so b is maybe

c is wrong

and d a physical change, will technically yes and no there should have not been a smell afterwards so the only really choose that makes sense here is A

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Determine the velocity of a 55-kg skier whose kinetic energy is 8900 J

goldfiish [28.3K]

Answer:

Kinetic energy = 1/2mv^2

8900 j =1/2*55*v^2

v^2=8900*2/55

v^2=323.6

v=17.98

Explanation:

4 0

2 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

Consider the atom with the chemical symbol Ru.

Vinil7 [7]

Answer: 97

Explanation:

The element Ru is Ruthenium which has 44 protons.

In this case, our atom is an isotope with 53 neutrons.

With this information, we can use the mass number formula to find this atom's mass number.

mass number = protons + neutrons

mass number = 44 protons + 53 neutrons

mass number = 97

4 0

1 year ago

Compare and contrast physical changes with chemical changes.

Alekssandra [29.7K]

There are several differences between a physical and chemical change in matter or substances. A physical change in a substance doesn't change what the substance is. In a chemical change where there is a chemical reaction, a new substance is formed and energy is either given off or absorbed.

3 0

3 years ago

is the approximate relation celsius = 1/2 fahrenheit a better approximation at higher or lower temperatures?

kirill115 [55]

Answer : The approximate relation Celsius = 1/2 Fahrenheit is a better approximation at higher temperatures

Explanation :

The formula for Celsius to Fahrenheit conversion is

$C = (F-32) \times \frac{5}{9}$

At lower temperature the value that needs to be subtracted (32) is large enough as a result the approximation "celsius = 1/2 fahrenheit " does not seem valid.

For example, 50 F is 10°C.

$(50 - 32) \times \frac{5}{9} = 10 C$

This is almost 1/5 of Fahrenheit temperature.

But at higher temperatures , the value becomes insignificant and also the ratio 5/9 tend to be equal to 0.5.

For example, 2000 F is 1093°C

$(2000 - 32) \times \frac{5}{9} = 1093 C$

This is almost half of Fahrenheit temperature.

Therefore , the approximate relation Celsius = 1/2 Fahrenheit is a better approximation at higher temperatures

8 0

3 years ago