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Anarel [89]
3 years ago
14

WHAT IS NORMAL MUSCLE PH

Chemistry
1 answer:
larisa [96]3 years ago
8 0
The skeletal muscle ph is typically 7.15
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Frank has a sample of steel that weighs 80 grams. If the density of his sample of steel is 8 g/cm3 ,what is the samples volume?
kirill115 [55]

Answer: 10 cm3

Explanation:

Mass of sample of steel (m) = 80 grams

density of sample of steel (p) = 8 g/cm3 samples volume (v) = ?

Since density is obtained by dividing mass by volume, the density of the steel sample is expressed as:

density = mass / volume

p = m / v

make v the subject formula

v = m / p

v = 80 grams / 8 g/cm3

v = 10 cm3

Thus, the samples volume is 10 cm3

4 0
3 years ago
What information does the first quantum number of an electron give?
bazaltina [42]

Answer:

с

Explanation:

the first quantum number of an electron gives the information about the energy level the electron is in

4 0
2 years ago
A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M Na
s344n2d4d5 [400]

Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f:  after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

Normality=molarity \times n-factor

<em>but in case of NaOH and HCl n-factor is 1 for each.</em>

hence

normality=molarity;

At equivalence point:  N_1V_1=N_2V_2

Before equivalence point : N_1V_1>N_2V_2

After the equivalence point: N_1V_1

N_1V_1=100\times1=100

case a:  5.00 mL of 1.00 M NaOH

N_2V_2=5\times1=5

N_1V_1>N_2V_2 hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

N_2V_2=100\times1=100

N_1V_1=N_2V_2 hence it is equivalence point

case c:  10.0 mL of 1.00 M NaOH

N_2V_2=10\times1=10

N_1V_1>N_2V_2 hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

N_2V_2=150\times1=150

N_1V_1 hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

N_2V_2=50\times1=50

N_1V_1>N_2V_2 hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

N_2V_2=200\times1=200

N_1V_1 hence it is after the eqivalence point

7 0
3 years ago
What's 2.5 meters to millimeters ????
irga5000 [103]

answer

2500

because it's mm

and you just have to add 2 more place values

4 0
2 years ago
Read 2 more answers
EXTRA CREDIT. The rate at which an object moves is its speed. If a horse
Irina18 [472]
Average speed is total distance divided by total time.

Total distance is 50 m + 150 m + 300 m = 500 m
Total time is 68 s + 35 s + 22 s = 125 s

Average speed is 500 m / 125 s = 4 m/s
6 0
3 years ago
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