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1 year ago

If an object goes through a chemical reaction but does not change in mass that object has

Chemistry

1 answer:

eimsori [14]1 year ago

6 0

Answer:

The law of conservation of mass

Explanation:

The law of conservation of mass

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In aqueous solution, hypobromite ion, BrO-, reacts to produce bromate ion, BrO3 -, and bromide ion, Br-, according to the follow

maw [93]

Answer:

22.73s

Explanation:

The reaction is a second order reaction, we know this by observing the unit of the slope.

rate constant = k = 0.056 M-1s-1

the initial concentration of BrO- [A]o = 0.80 M

time = ?

Final concentration [A]t= one-half of 0.80 M = 0.40M

1 / [A]t = kt + 1 / [A]o

1 / 0.40 = 0.056 * t + 1 / 0.80

t = (2.5 - 1.25) / 0.056

t = 22.73s

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2 years ago

Predict what would be formed

Anarel [89]

Answer:

Explanation:

T6tgbv. 55678 r4fyx a

8 0

2 years ago

: If you know that a ∝ b and a ∝ c, then you can also say that a ∝ bc, or the product of b and c. Take the above three proportio

aleksley [76]

Answer:

V ∝ abc

Explanation:

This task is a joint variation task involving only direct proportionality:

Direct variation is one in which two variables are in direct proportionality to each other. This means that as one increases, the other variable also increases and vice - versa.

Joint variation is one in which one variable is dependent on two or more variables and varies directly as each of them.

In this exercise:

If a ∝ b and a ∝ c, then a ∝ bc

Taking the above three proportionalities,

V ∝ a ∝ b ∝ c

V ∝ a ∝ bc

V ∝ abc

3 0

2 years ago

9. If 28.56 g of K2O is produced when 25.00 g K is reactedaccording to the following equation, what is the percent yieldof the r

Mumz [18]

Answer

b. 95%

Explanation

Given:

Mass of K₂O produced (actual yield) = 28.56 g

Mass of K that reacted = 25.00 g

Equation: 4K(s) + O₂(g) → 2K₂0(s)

What to find:

The percent yield of K₂O.

Step-by-step solution:

The first step is to calculate the theoretical yield of K₂O produced.

From the balanced equation, 4 mol K produced 2 mol K₂O

Molar mass of K₂O = 94.20 g/mol)

Molar mass of K = 39.10 g/mol)

This means 4 mol x 39.10 g/mol = 156.40 g K produced 2 mol x 94.20 g/mol = 188.40 g K₂O

So 25.00 g K will produce:

$\frac{25.00\text{ }g\text{ }K\times188.40\text{ }g\text{ }K₂O}{156.40\text{ }g\text{ }K}=30.1151\text{ }g\text{ }K₂O$

Actual yield of K₂O = 28.56 g

Theoretical yield of k₂O = 30.1151 g

The percent yield for the reaction can now be calculated using the formula below:

$\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\ \\ Percent\text{ }yield=\frac{28.56\text{ }g}{30.1151\text{ }g}\times100\% \\ \\ Percent\text{ }yield=0.9484\times100\% \\ \\ Percent\text{ }yield=94.84\%\approx95\% \end{gathered}$

Therefore, the percent yield for the reaction is 95%.

3 0

1 year ago

When 4.50 L of hydrogen gas react with an excess of nitrogen gas at standard temperature and pressure, how many liters of ammoni

VARVARA [1.3K]

Answer:

3.0 L of NH₃

Solution:

The equation is as follow,

N₂ + 3 H₂ → 2 NH₃

According to equation,

67.2 L (3 mole) H₂ at STP produces = 44.8 L (3 mole) of NH₃

So,

4.50 L of H₂ will produce = X L of NH₃

Solving for X,

X = (4.50 L × 44.8 L) ÷ 67.2 L

X = 3.0 L of NH₃

5 0

3 years ago