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Alinara [238K]
4 years ago
7

17. What is the average atomic mass of the following isotopic mixture - 22.00% of 159.3 g/mole; 78.00% of

Chemistry
1 answer:
NeX [460]4 years ago
6 0
(0.22)*(159.3) + (0.78)*(161.2)
34.046 + 125.736 = 159.8 g/mole
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How many equivalents of pyruvate are needed to generate 1 equivalent of glucose?
kodGreya [7K]
Gluconeogenesis is the process by which the body produce glucose from non-carbohydrate substrates such as pyruvate. To produce glucose from pyruvate, two moleucules of pyruvate is needed. The reaction for the gluconeogenesis reaction is as follow:
2 pyruvate + 4 ATP + 2 GTP + 2 NADH = Glucose + 4 ADP + 2 GDP + 2 NAD + 6Pi. 

4 0
3 years ago
what is the approximate ph at the equivalence point of a weak acid-strong base titration if 25 ml of aqueous formic acid require
vivado [14]

Answer:

pH at equivalence point is 8.52

Explanation:

HCOOH+NaOH\rightarrow HCOO^{-}Na^{+}+H_{2}O

1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of HCOO^{-}

So, moles of NaOH used to reach equivalence point equal to number of moles HCOO^{-} produced at equivalence point.

As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.

So, moles of HCOO^{-} produced = \frac{29.80\times 0.3567}{1000}moles=0.01063moles

Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL

So, at equivalence point concentration of HCOO^{-} = \frac{0.01063\times 1000}{54.80}M=0.1940M

At equivalence point, pH depends upon hydrolysis of HCOO^{-}. So, we have to construct an ICE table.

HCOO^{-}+H_{2}O\rightleftharpoons HCOOH+OH^{-}

I: 0.1940                                   0                 0

C: -x                                          +x               +x

E: 0.1940-x                                x                  x

So, \frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}

species inside third bracket represent equilibrium concentrations

So, \frac{x^{2}}{0.1940-x}=5.56\times 10^{-11}

or,x^{2}+(5.56\times 10^{-11}\times x)-(1.079\times 10^{-11})=0

So, x=\frac{-(5.56\times 10^{-11})+\sqrt{(5.56\times 10^{-11})^{2}+(4\times 1.079\times 10^{-11})}}{2}

So, x=3.285\times 10^{-6}M

So, pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52

5 0
3 years ago
A 35.0 g piece of metal wire is heated, and the temperature changes from 21 degrees Celsius to 52 degrees Celsius. The amount of
Elina [12.6K]

Answer:

1.23 j/g. °C

Explanation:

Given data:

Mass of metal = 35.0 g

Initial temperature = 21 °C

Final temperature = 52°C

Amount of heat absorbed = 320 cal  (320 ×4.184 = 1338.88 j)

Specific heat capacity of metal = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 52°C -  21 °C

ΔT = 31°C

1338.88 j= 35 g ×c× 31°C

1338.88 j= 1085 g.°C ×c

1338.88 j/1085 g.°C = c

1.23 j/g. °C = c

5 0
3 years ago
For this ionic compound, what would be the name of the anion? Ca(NO3)2
Ne4ueva [31]
Idk if it's correct but I guess it's calcite (c)
4 0
4 years ago
Read 2 more answers
OK OK ONE MORE TRY
babunello [35]
Your question can be found online! Just draw one of these diagrams

7 0
3 years ago
Read 2 more answers
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