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3 years ago

Sunny makes a poster showing the levels of

Chemistry

2 answers:

ZanzabumX [31]3 years ago

8 0

Answer:

it is the 4th and 5th one

Explanation:

i did it

Vladimir [108]3 years ago

5 0

Answer:

3rd and 5th!

Explanation:

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Let the simulator run for 30 seconds. What happens to the size of the ice?

Shkiper50 [21]

10 is your answer hope it helps you.......... Xd

<h2>Answer : 10 </h2>

7 0

3 years ago

Based on Chromium's position on the periodic table, which statement describes the element

natali 33 [55]

Answer:

C. chromium is a metal that is less reactive than sodium.

Explanation:

Hello.

Given the options:

A. chromium is a nonmetal and therefore a good conductor of heat and electricity .

B. chromium is a metal that is more reactive than potassium .

C. chromium is a metal that is less reactive than sodium .

D. chromium is a noble gas that is not reactive.

In this case, since chromium is in period 4 group VIB we infer it is a transition metal which slightly reacts with acids and poorly reacts with oxygen and other oxidizing substances. Thus, in comparison with both sodium and potassium which are highly reactive even with water as they get on fire, we can say that it is less reactive than both potassium and sodium, therefore, answer is: C. chromium is a metal that is less reactive than sodium.

Best regards.

7 0

3 years ago

What is the concentration of a saturated solution of SrSO4? SrSO4 has a Ksp of 3.2 x 10–7

Andrej [43]

SrSo4 = Sr(2+) + SO4(2-)

Let’s say that the initial concentration of SrSo4 was 1. ( or we have 1 mole of this reagent).

When The reaction occurs part of SrSo4is dissociated. And we get X mole Sr(2+) and So4(2-).
Ksp=[Sr(2+)]*[SO4(2-)]
X^2=3.2*10^-7
X=5.6*10^-4

5 0

3 years ago

1. What type of battery was made by alternating layers of silver, zinc and blotting paper soaked in salt?

Sunny_sXe [5.5K]

Answer:

voltaic piles

Explanation:

4 0

3 years ago

Read 2 more answers

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

3 0

3 years ago