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inysia [295]
3 years ago
11

A substance that dissolves another substance is a _____________________

Chemistry
2 answers:
Scorpion4ik [409]3 years ago
7 0

Answer:

b it is easy its b

Explanation:

balandron [24]3 years ago
5 0

Answer:

B

Explanation:

You might be interested in
A gas occupies 56 L at 73°C. What volume will the gas occupy if the temp. cools to 0°C?
vova2212 [387]

Answer:

44.2 L

Explanation:

Use Charles Law:

\frac{V1}{T1} =\frac{V2}{T2}

We have all the values except for V₂; this is what we're solving for. Input the values:

\frac{56 L}{346K} =\frac{V2}{273K}   -  make sure that your temperature is in Kelvin

From here, we need to get V₂ by itself. To do this, multiply by 273 on both sides:

\frac{56*273}{346} = V2

Therefore, V₂ = 44.2 L

It's also helpful to know that temperature and volume are linearly related. So, when temperature drops, so will volume and vice versa.

7 0
3 years ago
The temperature of a city during a week was 35° C, 36°C, 34°C, 38°C, 40°C, 39°C and 44°C. What was the average daily temperature
Zigmanuir [339]

Answer:38°

Explanation:

8 0
3 years ago
Identify the limiting reactant when 32. 0 g hydrogen is allowed to react with 16. 0 g oxygen
Mkey [24]

Answer:

Oxygen is limiting reactant

Explanation:

2 H2  +   O2  ======> 2 H2 O

from this equation (and periodic table) you can see that

  4 gm of H combine with 32 gm O2  

     H / O  =  4/32 = 1/8

       32 /16    =  2/1    shows O is limiter

         for 32 gm H you will need 256 gm O   and you only have 16 gm

3 0
2 years ago
A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
Igoryamba

Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

          V_{2} = 8.34 L,    V_{1} = 2.67 L

Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

    = -2818.68 J

Hence, work for path A is -2818.68 J.

  • For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

                   = -1 atm \times (8.34 L - 2.67 L)  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = -\frac{101.33 J}{1 atm L} \times 5.67 atm L

            = -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

3 0
3 years ago
What can the presence of suffixes such as -ite,-ate,or -de in a chemical name indicate?
laila [671]

Answer: Most polyatomic ions are -ate, related ions that have 1 fewer oxygen atom are -ite (must have at least 1 oxygen tho) and nonmetals are ide.

Explanation:

3 0
1 year ago
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