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Marrrta [24]
3 years ago
15

Which has a higher average speed of molecules 95 degrees Celsius or 95 degrees Fahrenheit ​

Chemistry
2 answers:
denis23 [38]3 years ago
7 0
95 celsius lol ur questions r so easy
shutvik [7]3 years ago
6 0

Answer:

95(ninety five) degrees Celsius

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if you measure the mass of a model car to be 230 grams. the actual mass is 218 grams. what is your percent error.
Roman55 [17]
27.6% is the correct answer 
6 0
3 years ago
10 points
kaheart [24]

Answer:

False.

Explanation:

I am not completely sure, but I don't believe the earth is getting larger.

8 0
3 years ago
A 72.0 mL aliquot of a 1.40 M solution is diluted to a total volume of 248 mL. A 124 mL portion of that solution is diluted by a
Aliun [14]

Answer: 0.20 M

Explanation:

According to the dilution law,

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 1.40 M

V_1 = volume of stock solution = 72.0 ml

M_2 = molarity of diluted solution = m

V_2 = volume of diluted solution = 248 ml

1.40\times 72.0=m\times 248

m=0.41M

Now 124 mL portion of this prepared solution is diluted by adding 133 mL of water.

According to the dilution law,

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 0.41 M

V_1 = volume of stock solution = 124 ml

M_2 = molarity of diluted solution = m

V_2 = volume of diluted solution = (124 +133) ml = 257 ml

0.41\times 124=m\times 257

m=0.20M

Thus the final concentration of the solution is 0.20 M.

8 0
3 years ago
Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib
Dvinal [7]

<u>Answer:</u> The value of K_c for the final reaction is 7.16\times 10^{25}

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u>  H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1

<u>Equation 2:</u>  HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2

The net equation follows:

S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

K_c=\frac{1}{K_1}\times \frac{1}{K_2}

We are given:

K_1=9.57\times 10^{-8}

K_2=1.46\times 10^{-19}

Putting values in above equation, we get:

K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}

Hence, the value of K_c for the final reaction is 7.16\times 10^{25}

5 0
3 years ago
2. How many grams of water can be warmed from 25.0°C to 37.0°C by the addition of 8,064 calories?
sertanlavr [38]

Answer:

672 g

Explanation:

We can calculate the mass of water that can be warmed from 25.0°C to 37.0°C by the addition of 8,064 calories using the following expression.

Q = c \times m \times \Delta T

where,

c: specific heat of the water

m: mass

ΔT: change in the temperature

m = \frac{Q}{c \times \Delta T  }  = \frac{8,064cal}{(1cal/g. \° C) \times (37.0 \° C - 25.0 \° C)  } = 672 g

The mass of water that can be warmed under these conditions is 672 grams.

5 0
3 years ago
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