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3 years ago

12

You place a balloon in a closed chamber at STP. When the balloon doubles in size, what happens to the chamber pressure? A.) The

pressure stays the same B.) The pressure doubles C.) The pressure reduced by one half

2 answers:

Katarina [22]3 years ago

8 0

It decrease's so C is correct.

larisa86 [58]3 years ago

4 0

If you increase the volume the pressure will decrease. so the best answer is C

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A 19.0 g piece of metal is heated to 99.0 °C and then placed in 150 mL of water at 21°C. The temperature of the water rose to 23

Anna35 [415]

Answer:

$0.8696\ \text{J/g}^{\circ}\text{C}$

Explanation:

$m_m$ = Mass of metal = 19 g

$c_m$ = Specific heat of the metal

$\Delta T_m$ = Temperature difference of the metal = $99-23=76^{\circ}\text{C}$

V = Volume of water = 150 mL = $150\ \text{cm}^3$

$\rho$ = Density of water = $1\ \text{g/cm}^3$

$c_w$ = Specific heat of the water = 4.186 J/g°C

$\Delta T_w$ = Temperature difference of the water = $23-21=2^{\circ}\text{C}$

Mass of water

$m_w= \rho V\\\Rightarrow m_w=1\times 150\\\Rightarrow m_w=150\ \text{g}$

Heat lost will be equal to the heat gained so we get

$m_mc_m\Delta T_m=m_wc_w\Delta T_w\\\Rightarrow c_m=\dfrac{m_wc_w\Delta T_w}{m_m\Delta T_m}\\\Rightarrow c_m=\dfrac{150\times 4.186\times 2}{19\times 76}\\\Rightarrow c_m=0.8696\ \text{J/g}^{\circ}\text{C}$

The specific heat of the metal is $0.8696\ \text{J/g}^{\circ}\text{C}$ .

4 0

3 years ago

What is the Molar mass of Na3P

Sedaia [141]

Answer:

99.943071 g/mol

Explanation:

22.98977*3+30.973761

6 0

3 years ago

A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an

erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

7 0

3 years ago

What is the name of MnO? Explain how you determined the bond type and the steps you used to determine the naming convention for

xeze [42]

Answer:

Magnesiumoxide

Explanation:

I don't know

5 0

3 years ago

Nault 25000L 250 our = 25000L 15040 8. How many grams of CaCl2 are needed to make 150.0 mL of a 0.500 M CF solution? (Note: CaCl

yuradex [85]

Answer:

You need 8,324 g of CaCl₂ yo make this solution

Explanation:

Molarity is a way to express concentration in a solution, in units of moles of solute per liter of solution.

To know the grams of CaCl₂ it is necessary to know, first, the moles of this substance with the desired volume and concentration , thus:

0,1500 L × $\frac{0,500 mol}{L}$ = 0,075 CaCl₂ moles

Now, with the molar mass of CaCl₂ you will obtain the necessary grams, thus:

0,075 CaCl₂ moles × $\frac{110,98 g}{mol}$ = 8,324 g of CaCl₂

So, you need <em>8,324 g of CaCl₂</em> to make 150,0 mL of a 0,500M solution

I hope it helps!

3 0

3 years ago