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3 years ago

13

You put one ice cube on a room temperature rock and another ice cube on a room temperature piece of foil. Which ice cube will me

lt faster? Why?

1 answer:

aksik [14]3 years ago

5 0

Answer:

a ice cube on a room temperature rock

Explanation:

because the air temperature is warmer than the freezers which means the ice particle gather heat energy from warmer air.

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What evidence supports a conservation law? 6 CO2 → C6H12O6 6 O2 6 CO2 6 H2O light → C6H12O6 6 O2 6 H2O light → C6H12O6 6 O2 6 CO

docker41 [41]

The law of conservation has been stated that the mass and energy has neither be created nor destroyed in a chemical reaction.

The law of conservation has been evident when there has been an equal number of atoms of each element in the chemical reaction.

<h3>Conservation law</h3><h3 />

The given equation has been assessed as follows:

$\rm 6\;CO_2\;\rightarrow\;C_6H_1_2O_6$

The reactant has absence of hydrogen, while hydrogen has been present in the product. Thus, the reaction will not follow the law of conservation.

$\rm 6\;O_2\;+\;+\;6\;CO_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6$

The number of atoms of each reactant has been different on the product and the reactant side. Thus, the reaction will not follow the law of conservation.

$\rm 6\;O_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6$

The reactant has the presence of carbon, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

$\rm 6\;O_2\;+\;6\;CO_2\;\rightarrow\;3\;C_6H_1_2O_6\;+\;3\;O_2$

The product has the presence of hydrogen, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

Learn more about conservation law, here:

brainly.com/question/2175724

4 0

3 years ago

What is the complete ionic equation for H2SO4(aq) + Cal2(aq) -> CaSO4(s) + 2HI(aq)?

eimsori [14]

2H(+) + SO4(2-) + Ca(2+) + 2I(-) -> CaSO4(s) + 2H(+) + 2I(-)
The signs in brackets are the subscripts for the charge of the ion. This is the complete ionic equation. The net ionic equation is:
Ca(2+) + SO4(2-) -> CaSO4

6 0

3 years ago

Read 2 more answers

Tin(ii) fluoride (snf2) is often added to toothpaste as an ingredient to prevent tooth decay. what is the mass of f in grams in

a_sh-v [17]

From the periodic table:
molar mass of Sn = 118.71 grams
molar mass of F = 18.99 grams

This means that:
molar mass of SnF2 = 118.71 + 2(18.99) = 156.69 grams

Therefore, 156.69 grams of SnF2 contains 37.98 grams of F. To know the amount of F in 36.5 grams of the compound, we will simply do a cross multiplication as follows:
mass of F = (36.6 x 37.98) / 156.69 = 8.847 grams

3 0

3 years ago

A chemical change in which a single compound breaks down into two or more products is which of the following?

Blizzard [7]

Answer:

...

Explanation:

6 0

3 years ago

Read 2 more answers

Be sure to answer all parts. Express the rate of reaction in terms of the change in concentration of each of the reactants and p

Tanzania [10]

Answer : The [H] is increasing at the rate of 0.36 mol/L.s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}$

$\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}$

$\text{Rate of disappearance of }F=-\frac{d[F]}{dt}$

$\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}$

$\text{Rate of formation of }H=+\frac{d[H]}{dt}$

$\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}$

Given:

$-\frac{d[D]}{dt}=0.18mol/L.s$

As,

$-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s$

and,

$+\frac{d[H]}{dt}=2\times 0.18mol/L.s$

$+\frac{d[H]}{dt}=0.36mol/L.s$

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

5 0

3 years ago