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Anvisha [2.4K]
3 years ago
8

Economics is about the production of goods and services which requires that allocation of what

Chemistry
1 answer:
yuradex [85]3 years ago
6 0

It requires the allocation of resources.

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hydrogen and oxygene react to form water l. How many molecules of O2 are required to produce 0.6 of H20​
user100 [1]

Answer:

It is required 0.3 molecules of O2 to produce 0.6 molecules of H2O

Explanation:

The water´s formation reaction is: 2H2 + O2 --> 2H2O

This reaction shows the molecular O2:H2O ratio: 1:2.

Then, if we want to know how many molecules of O2 are required to produce 0.6 of H20​, it is necessary calculate as it showed next:

0.6 molecules H2O * (1 molecula O2 / 2 molecules H2O)= 0.3 molecule O2

8 0
3 years ago
How much salt (NaCl) is carried by a river flowing at 30.0 m3/s and containing 50.0 mg/L of salt
denis23 [38]

Answer:

129,600kg/day

Explanation:

The river is flowing at 30.0

1 m³ is equivalent to 1000L

flowrate of river = 30*1000 =30,000L/s

Convert L/s to litre per day by multiplying by 24*60*60

flowrate of river = 30,000 * 24*60*60 L/day

                    = 2,592,000,000L/day

if the river contains 50mg of salt  in 1L of solution

lets find how many mg of salt Y is contained in 2,592,000,000L/day

by cross multiplying we have

Y=\frac{2592000000*50}{1}

Y= 129,600,000,000 mg/day

convert this value to kg/day by dividing by 1 million

Y= 129,600,000,000/1000000

Y= 129,600kg/day

3 0
3 years ago
The ratio between the force of sliding friction and the normal force of an object is called _____.
Kitty [74]

Answer: The correct answer is C.

Explanation: Frictional force is the contact force which resists the motion of one surface on another.

Mathematically,

f=\mu N

where,

f = frictional force

\mu = Coefficient of friction

N = normal force acting on a body

We need to find the ratio of force of sliding friction ( which is a type of frictional force) and the normal force acting on an object. This will be equal to coefficient of friction.

\mu=\frac{f}{N}

4 0
3 years ago
The second-order rate constant for the following gas-phase reaction is 0.041 1/MLaTeX: \cdotâs. We start with 0.438 mol C2F4 in
pantera1 [17]

Answer:

134.8 seconds is the half-life (in seconds) of the reaction for the initial C_2F_4 concentration

Explanation:

Half life for second order kinetics is given by:

t_{\frac{1}{2}=\frac{1}{k\times a_0}

Integrated rate law for second order kinetics is given by:

\frac{1}{a}=kt+\frac{1}{a_0}

t_{\frac{1}{2} = half life

k = rate constant

a_0 = initial concentration

a = Final concentration of reactant after time t

We have :

C_2F_4 \longrightarrow \frac{1}{2} C_4F_8

Initial concentration of C_2F_4=[a_o]=\frac{0.438 mol}{2.42 L}=0.1810 mol/L

Rate constant = k = 0.041 M^{-1} s^{-1}

t_{\frac{1}{2}=\frac{1}{k\times a_0}

=\frac{1}{0.041 M^{-1} s^{-1}\times 0.1810 mol/L}

t_{1/2}=134.8 s

134.8 seconds is the half-life (in seconds) of the reaction for the initial C_2F_4 concentration

3 0
3 years ago
What is the formula of titanium(IV) bromide?
Anna [14]

Answer:

\boxed{\text{SnBr}_{4}}

Explanation:

The name tells you that this is a binary compound (contains two elements).

It contains a metal and a nonmetal, so it is a binary ionic compound. The general rule is:

Name of compound = name of metal  name of ion (two words)

Name of metal = tin(IV), so the tin ion has a charge of 4+

Name of ion = bromide. Br is in Group 17, so bromide ion has charge of 1-.

\rm Sn$^{4+}$ + 4Br$^{-} \longrightarrow \,$ SnBr$_{4}$\\\\\text{The formula is} \boxed{\textbf{SnBr}_{4}}

3 0
3 years ago
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