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2 years ago

How many grams of caffeine should we prepare in a 10 mL standard of 3 mm

Chemistry

2 answers:

Brut [27]2 years ago

8 0

Answer:

if it is espresso(1 shot), the coffee content to be considered is: (77/44)*(10/3)= 5.833= 0.005833 grams

Explanation:

gregori [183]2 years ago

8 0

Answer:

d) 5 grams

Explanation:

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Which of the following experiments would best test viscosity?

Juli2301 [7.4K]

C. pouring honey on a plate so the density of thickness and stickyness would
be the highest

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3 years ago

How does nuclear fission of Uranium - 234 result in electricity being generated?

nevsk [136]

Answer:

See explanation

Explanation:

The use of Uranium - 234 to generate electricity depends on a fission reaction. The uranium nuclide is bombarded by fast moving neutrons leading to a chain reaction. Control rods and moderators are used to keep the nuclear reaction under control.

As the nuclear reaction proceeds, heat is generated and steam is consequently produced. This steam is used to turn a turbine and electricity is thereby generated.

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3 years ago

Hello can you please help me to solve above fill in the blank questions

LiRa [457]

Answer:

C2H5OH is a formula of ethyl alchol or ethanol.

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Explanation:

i hope this will help you :)

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3 years ago

Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec

Svetach [21]

The overall reaction is given by:

$Br_{2}(g) + 2 NO(g) \rightarrow 2 NOBr(g)$

The fast step reaction is given as:

$NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})$

The slow step reaction is given as:

$NOBr_{2}(g) + NO(g) \rightarrow 2 NOBr(g)$ (slow step $k_{2}$ )

Now, the expression for the rate of reaction of fast reaction is:

$r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]$

The expression for the rate of reaction of slow reaction is:

$r_{2}=k_{2}[NOBr_{2}] [NO]$

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as $[NOBr_{2}]$ takes place in this reaction.

The expression of rate of formation is:

$\frac{d(NOBr)}{dt}=r_{2}$

= $k_{2}[NOBr_{2}][NO]$ (1)

Now, consider that the fast step is always is in equilibrium. Therefore, $r_{1}=0$

$k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]$

$[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]$

Substitute the value of $[NOBr_{2}]$ in equation (1), we get:

$\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]$

= $k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]$

= $\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]$

Thus, rate law of formation of $NOBr$ in terms of reactants is given by $\frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]$ .

4 0

3 years ago

What is the value of k, at 25 degrees Celsius

Dimas [21]

<h3>Answer:</h3>

298.15 K

<h3>Explanation:</h3>

W e are supposed to calculate the Value of K at 25°C

Assuming the value of K represent K, the question wants us to convert degree Celsius to Kelvin.

To convert degrees Celsius to kelvin scale, we use the relationship;
Kelvin (K) = Degrees Celsius + 273.15 ; 273.5 is a constant
That is, to convert temperature from °C to Kelvin we add a constant of 273.15 to the °C given.

In this case;

Temperature is 273.15 °c

Thus, to Kelvin scale temperature will be;

= 25°C + 273.15

= 298.15 K

Therefore, the value of K, at 25°C is 298.15 K

7 0

3 years ago