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2 years ago

Zinc + Hydrogen Chloride What is proof of a chemical reaction?

Chemistry

2 answers:

maria [59]2 years ago

7 0

Answer:

zncl2. . . . . . . . . . . . . . . .

alexira [117]2 years ago

5 0

Answer:

produces bubbles and hydrogen gas

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Which energy conservation occurs when you start running

Lady_Fox [76]

Answer:

Kinetic

Explanation:

Kinetic energy kicks in after potienal builds up energy

4 0

2 years ago

Read 2 more answers

What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a

Vadim26 [7]

Answer:

$\boxed {\boxed {\sf 333 \ grams}}$

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

$Q= mc \Delta T$

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

ΔT= final temperature - initial temperature

The sample was heated from 58.8 degrees Celsius to 88.9 degrees Celsius.

ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

Q= 4500.0 J
c= 0.4494 J/g°C
ΔT = 30.1 °C

Substitute these values into the formula.

$4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)$

Multiply on the right side of the equation. The units of degrees Celsius cancel.

$4500.0 \ J = m (13.52694 J/g)$

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

$\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}$

The units of Joules cancel.

$\frac {4500.0 \ J }{13.52694 J/g}= m$

$332.6694729 \ g =m$

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

$333 \ g \approx m$

The mass of the sample of metal is approximately 333 grams.

8 0

2 years ago

If 26.2 mL of AgNO3 is needed to precipitate all the Cl− ions in a 0.785-mg sample of KCl (forming AgCl), what is the molarity o

sweet-ann [11.9K]

Answer:

The molarity of the $AgNO_3$ solution is $4.02 \times 10^4 M$

Explanation:

The Balanced chemical equation is

$1AgNO_3 (aq) +1KCl (aq) > 1 AgCl (s)+1KNO_3 (aq)$

Mole ratio of $AgNO_3$ : KCl is 1 : 1

So moles $AgNO_3$ = moles KCl

$Moles KCl = \frac {mass}{molarmass}$

$= \frac {0.785 mg}{(39.1+35.5 g per mol)}$

$= \frac {0.000785 g}{74.6 g per mol}$

$= 0. 0000105 mol KCl$

$= 0.0000105 mol AgNO_3$

So Molarity

$= \frac {moles of solute}{(volume of solution in L)}$

$= \frac {0.0000105 mol}{26.2 mL}$

$=\frac {0.0000105 mol}{0.0262 L}$

= 0.000402M or mol/L is the Answer

(Or) $4.02 \times 10^4 M$ is the Answer

6 0

2 years ago

Determine the formula weight of BeCl₂.

tia_tia [17]

Be= Beryllium= weight 9.01
Cl= Chlorine= weight 35.45
Since there are 2 Chlorine, you have to add Chlorine twice.
9.01 + 35.45 + 35.45= 79.91

Answer: 79.91
(This answer does not include sig.figs)

4 0

3 years ago

Which of the following is an isotope of hydrogen

Lena [83]

Z=1 is the formula i would have to see the following lol

7 0

2 years ago