All categories

3 years ago

Match these items

2 answers:

4 0

The correct matches are as follows:

1.instantaneous combustion
G.burning

2.mass of substances before and after a reaction is the same
C.Law of Conservation of Matter

3.substances that combine
A.reactants

4. Yields or makes
B.arrow symbol

5.rapid oxidation
F.explosion

6.new substance
D.product

7.slow oxidation
E.rust

Hope this answers the question. Have a nice day.

Fiesta28 [93]3 years ago

4 0

The CORRECT answers are:

Explosion: Instantaneous Combustion

Law of Conservation of Matter : Mass of substances before and after a reaction is the same

Reactants: Substances that combine

Arrow Symbol: Yields

Burning: Rapid Oxidation

Product: New Substance

Rust: Slow Oxidation

Ths are ALL Correct! Enjoy!

You might be interested in

Nvmmmmmmmmmmmmmmmmmmmmmmmmm lol

Vlad1618 [11]

Answer:

coooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooool lol

Explanation:

7 0

3 years ago

Read 2 more answers

A fossilized animal that was killed by a volcanic eruption has 6.25 percent of its original amount of carbon-14 remaining; this

Art [367]

22 million years ago

7 0

3 years ago

Read 2 more answers

What does solubility mean?

jasenka [17]

Answer: the ability to be dissolved, especially in water.

Explanation: I think the answer you've picked is right

Hope this helps

8 0

3 years ago

If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution

Helen [10]

This is a straightforward dilution calculation that can be done using the equation

$M_1V_1=M_2V_2$

where M₁ and M₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and V₁ and V₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (M₁) and the initial (V₁) and final (V₂) volumes, and we want to find the final concentration (M₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for M₂:

$M_2=\frac{M_1V_1}{V_2}.$

Substituting in our values, we get

$\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].$

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

5 0

2 years ago

Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs

saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of $Ag^+$ when CuBr just begins to precipitate is, $1.34\times 10^{-6}M$

Percent of $Ag^+$ remains is, 0.0018 %

Explanation :

$K_{sp}$ for CuBr is $4.2\times 10^{-8}$

$K_{sp}$ for AgBr is $7.7\times 10^{-13}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

$CuBr\rightleftharpoons Cu^++Br^-$