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jenyasd209 [6]
3 years ago
12

How many prime numbers are there between 0 and 75​

Chemistry
2 answers:
slamgirl [31]3 years ago
8 0

Answer:

There are 21 prime numbers between 0 and 75.

Explanation:

2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73

Paraphin [41]3 years ago
3 0

Answer:

21

Explanation:

There are 21 prime numbers between 0 and 75​.

These include:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73.

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A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

5 0
3 years ago
Si 100 monedas tienen una masa total de 226.13g. Cual es la masa de una moneda
galben [10]

The mass of a coin : 2.2613 g

<h3>Further explanation  </h3>

Algebraic expressions in mathematics are a combination of coefficients, numbers, variables, constants and arithmetic operations such as addition, subtraction, multiplication, and division.  

The main composition of algebraic expressions are:  

  • 1. phrases  

algebraic forms separated by arithmetic operations  

consists of one phrase (monomial) to many phrases (polynomial)  

  • 2. variables  

is a value that can be changed, can be in the form of letters, for example, x, y, a, b, etc.  

  • 3. constants  

is a fixed value, can be a number  

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<em>If 100 coins have a total mass of 226.13g. What is the mass of a coin</em>

We can make this statement in the form of an algebraic equation :

100 coins=226.13 g

then the mass of a coin :

\tt \dfrac{1}{100}\times 226.13~g=2.2613~g

5 0
3 years ago
Dalton's law of partial pressures states that the total pressure exerted by a mixture of gases is the sum of the pressures exert
Damm [24]

Answer: The given statement is true.

Explanation:

According to the Dalton's law, total pressure of a mixture of gases that do not react with each other is equal to the partial pressure exerted by each gas.

The relationship is as follows.

          p_{total} = \sum_{i=1}^{n} p_{i}

or,        p_{total} = p_{1} + p_{2} + p_{3} + p_{4} + ......... + p_{n}

where,  p_{1}, p_{2}, p_{3} ....... = partial pressure of individual gases present in the mixture

Also, relation between partial pressure and mole fraction is as follows.

                 p_{i} = p_{total} \times x_{i}

where,      x_{i} = mole fraction

Thus, we can conclude that the statement Dalton's law of partial pressures states that the total pressure exerted by a mixture of gases is the sum of the pressures exerted independently by each gas in the mixture, is true.              

5 0
3 years ago
1. The equation for the reaction between zinc and aqueous copper(II) sulfate is shown.
Fed [463]

Answer:

The oxidation state of the reducing agent has changed from 0 to +2.

Explanation:

reducing agent is anything that loses electron or gains oxygen

in this case, zinc

3 0
3 years ago
What are the units of k in the following rate law?
Ludmilka [50]

Answer:

B. \frac{1}{M^{2} s }

Explanation:

The unit for rate is M/s while the unit for each molecule should be M. You can find the unit for k by putting the units for rate and the molecules into the equation

rate= k{X][Y]

M/s= k * M^{2} * M^{1}

k= (M/s) / (M^{3})

k= \frac{1}{M^{2} s }

You can also use this predetermined formula to solve this problem faster: k= \frac{M^{1-n} }{s }

Where n is the number of molecule. There are 3 molecule(2X and 1Y) so n=3, so

k= \frac{M^{1-n} }{s }

k= \frac{M^{1-3} }{s }= \frac{m^{-2}}{s}= \frac{1}{M^{2} s }

8 0
3 years ago
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