All categories

2 years ago

What is the chemical formula of a trisaccharide made of three bonded glucose molecules?

1 answer:

3 0

(6 C, 12 H, 6 O) x 3 - (H20) x 2 = 18 carbons, 32 hydrogens, 16 oxygens

the answer is C18H32O16
because you use 2 water molecules to form molecule...so you are losing 4 H molecules and 2 O molecules). the first 2 glucose molecule combine losing one water molecule; and the third combines to them, losing another water molecule.

You might be interested in

How is a series circuit different from a parallel circuit?

beks73 [17]

Answer:

C. A series circuit has only one loop and a parallel circuit has two or more loops for the current to flow through

Explanation:

A circuit that are made of one loop is called series circuit. On the other hand, the parallel circuit has at least two loops. The circuit type has nothing to do with open or closed circuit.

If any part of the series circuit got cut, the current will stop flowing since there is only one loop. A parallel circuit has more loop so the circuit might still work even if a part of the circuit got cut.

4 0

3 years ago

A piece of copper alloy with a mass of 85.0 g is heated from 30. °C to 45 °C. In the process, it absorbs 523 J of energy as heat

Ne4ueva [31]

Answer:

410.196 J/[kg*°C].

Explanation:

1) the equation of the energy is: E=c*m*(t₂-t₁), where E - energy (523 J), c - unknown specific heat of copper, m - mass of this copper [kg], t₂ - the final temperature, t₁ - initial temerature;

2) the specific heat of copper is:

$c=\frac{E}{m*(t_2-t_1)}; \ => \ c=\frac{523}{0.085*(45-30)}=\frac{523}{1.275}=410.196[\frac{J}{kg*C}].$

6 0

2 years ago

The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×

Anuta_ua [19.1K]

Answer:

$Kc=~1.49x10^3^4}$

Explanation:

We have the reactions:

A: $N_2_(_g_) + O_2_(_g_) 2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9$

Our target reaction is:

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)$

We have $NO_(_g_)$ as a reactive in the target reaction and $NO_(_g_)$ is present in A reaction but in the products side. So we have to flip reaction A.

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

Then if we add reactions A and B we can obtain the target reaction, so:

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9$

For the final Kc value, we have to keep in mind that when we have to add chemical reactions the total Kc value would be the multiplication of the Kc values in the previous reactions.

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}$

$Kc=~1.49x10^+^3^4}$

3 0

2 years ago

The radioisotope phosphorus-32 is used in tracers for measuring phosphorus uptake by plants. The half-life of phosphorus-32 is 1

Harman [31]

Answer:

54 days

Explanation:

We have to use the formula;

0.693/t1/2 =2.303/t log Ao/A

Where;

t1/2= half-life of phosphorus-32= 14.3 days

t= time taken for the activity to fall to 7.34% of its original value

Ao=initial activity of phosphorus-32

A= activity of phosphorus-32 after a time t

Note that;

A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)

Substituting values;

0.693/14.3 = 2.303/t log Ao/0.0734Ao

0.693/14.3 = 2.303/t log 1/0.0734

0.693/14.3 = 2.6/t

0.048=2.6/t

t= 2.6/0.048

t= 54 days

3 0

2 years ago

What does reaction b tell you about the type of molecule h2co3 is?

irinina [24]

H2o means water and co3 is co-signs

6 0

2 years ago