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GaryK [48]
2 years ago
14

Thủy phân m gam mantozo, sau một thời gian thu được dung dịch X. Khi cho dung dịch X tác dụng gần hết với dung dịch AgNO3 trong

NH3 thì thu được tối đa 194,4 gam Ag. Biết hiêu suất quá trình thủy phân là 80%. Giá trị gần nhất của m là :
Chemistry
1 answer:
Westkost [7]2 years ago
4 0

Answer:

I dont undarsatnd 2gebwhsanKM<dwkdwndwkjdwnfwkjdnfkwnfwkf

Explanation:

wnkf mnf wnmd

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Which of the following are acids by the Bronsted-Lowry Definition:
olga55 [171]

ANSWER IS (A)

EXPLANATION:

Bronsted-Lowry concept states that a substance is an acid if it can act as a H+ donor.

HCl in aqueous solution means that HCl is present in water, HCl + H2O --> H3O+ + Cl-.  This reaction will take place, the H+ from HCl will be donated to H2O. So, HCl is a bronsted-lowry acid by definition.


However, Methanol (CH3OH) its written that it is liquid, i.e. pure methanol, CH3OH(l).  It is both acidic as well as basic. when it is mixed with water then it behaves as an acid.

The last one ammonia in gas phase is also neutral because its not in water. if mixed in water it behaves as a base.



6 0
3 years ago
I NEED ANSWER ASAP Which of the following statements is true for a scientific theory? It is a well-tested hypothesis. It is veri
Digiron [165]

Answer:

A.) It is a well-tested hypothesis.

7 0
3 years ago
Read 2 more answers
How many moles are in 337 grams of tellurium?
MaRussiya [10]

about 43001 is it i think

3 0
3 years ago
Wendy is using a poorly calibrated electronic balance to measure the mass of a crucible. Her technique in making the measurement
Nataly [62]
B answer is b I repeat it’s b
4 0
3 years ago
A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

Molarity=\frac{moles}{\text{Volume of solution(L)}}

Moles of glucose = \frac{18.5 g}{180 g/mol}=0.1028 mol

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = \frac{0.1028 mol}{0.1 L}=1.028 mol/L

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = M_1=1.208 mol

Volume of the solution taken = V_1=30.0 mL

Molarity of the solution after dilution = M_2

Volume of the solution after dilution= V_2=0.500L = 500 mL

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

M_2=0.07248 mol/L

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

Moles of glucose = 0.07248 mol/L\times 0.1 L=0.007248 mol

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

4 0
3 years ago
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