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weqwewe [10]
3 years ago
13

1cal =———J A. 4.18 B. 5 C. 8 D. 3

Chemistry
1 answer:
stiv31 [10]3 years ago
4 0
The answer is A. In case you wanted to know, the actual answer is 4.184. If anything is unclear, just ask!
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What is the frequency of a wave that has 3 waves every five seconds?
Dmitry [639]

Answer:

Frequency, f = 0.6 Hz

Explanation:

We have,

Number of waves passing through a point are 3

Time for which the waves are passing is 5 seconds

It is required to find the frequency of a wave. The frequency of a wave is defined as the no of waves per unit time. So,

f=\dfrac{n}{t}\\\\f=\dfrac{3}{5}\\\\f=0.6\ Hz

So, the frequency of a wave is 0.6 Hz.

7 0
3 years ago
Calculate the number of grams of solute in 500.0 mL of 0.189 M KOH.
KIM [24]

Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

Solution : Given,

Volume of solution = 500 ml

Molarity of KOH solution = 0.189 M

Molar mass of KOH = 56 g/mole

Formula used :

Molarity=\frac{\text{Mass of KOH}\times 1000}{\text{Molar mass of KOH}\times \text{Volume of solution in ml}}

Now put all the given values in this formula, we get the mass of solute KOH.

0.189M=\frac{\text{Mass of KOH}\times 1000}{(56g/mole)\times (500ml)}

\text{Mass of KOH}=5.292g

Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

7 0
3 years ago
Read 2 more answers
What is the elevation of hachure line A?<br><br> 125 feet<br> 75 feet<br> 100 feet<br> 50 feet
mote1985 [20]
<span> elevation between index contours would be </span><span>125 feet</span>
7 0
3 years ago
Read 2 more answers
For the chemical reaction:
Kamila [148]

Answer:

Volume of ammonia produced = 398.7 dm³

Explanation:

Given data:

Volume of N₂ = 200 dm³

Pressure and temperature = standard

Volume of ammonia produced = ?

Solution:

Chemical equation:

N₂ + 3H₂     →      2NH₃

Number of moles of N₂:

PV = nRT

1 atm× 200 L = n× 0.0821 atm.L/mol.K × 273 K

n = 200 atm.L /22.41 atm.L/mol

n = 8.9 mol

Now we will compare the moles of ammonia and nitrogen.

               N₂          :         NH₃

                1            :           2

              8.9          :        2/1×8.9 = 17.8 mol

Volume of ammonia:

1 mole of any gas occupy 22.4 dm³ volume

17.8 mol ×22.4 dm³/1 mol = 398.7 dm³

7 0
3 years ago
The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.
madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

H_3BO_3\rightarrow H^+H_2BO_3^-

at t=0  cM              0             0

at eqm c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 4.4 M and \alpha = ?

K_a=5.8\times 10^{-10}

Putting in the values we get:

5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}

(\alpha)=0.000011

[H^+]=c\times \alpha

[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M

Also pH=-log[H^+]

pH=-log[4.8\times 10^{-5}]=4.3

Thus pH of a 4.4 M H_3BO_3 solution is 4.3

3 0
3 years ago
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