All categories

3 years ago

A

Chemistry

1 answer:

Luden [163]3 years ago

5 0

Answer:

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

$P_2 = \frac{P_1V_1}{V_2} \\$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

From the question we have

$P_2 = \frac{125\times3.50 }{2.75} = \frac{437.5}{2.75} \\ = 159.090909...$

We have the final answer as

Hope this helps you

You might be interested in

If the concentration of the HCl used in your titration was 1.00 M and you used 23.68 mL to reach the endpoint, calculate the con

drek231 [11]

Answer: The concentration of the OH-, CB = 0.473 M.

Explanation:

The balanced equation of reaction is:

2HCl + Ca(OH)2 ===> CaCl2 + 2H2O

Using titration equation of formula

CAVA/CBVB = NA/NB

Where NA is the number of mole of acid = 2 (from the balanced equation of reaction)

NB is the number of mole of base = 1 (from the balanced equation of reaction)

CA is the concentration of acid = 1M

CB is the concentration of base = to be calculated

VA is the volume of acid = 23.65 ml

VB is the volume of base = 25mL

Substituting

1×23.65/CB×25 = 2/1

Therefore CB =1×23.65×1/25×2

CB = 0.473 M.

6 0

3 years ago

Please help as fast as u can

pav-90 [236]

Answer:D

DDD

DDDDDDDDDD

Explanation:

6 0

3 years ago

Read 2 more answers

A gas under an initial pressure of 0.60 atm is compressed at constant temperature from 27 L to 3.0 L. The final pressure becomes

kherson [118]

The answer is.... : 5.4 atm

3 0

3 years ago

Read 2 more answers

If 4.5 moles of AlCl3 were made, how many moles of CuCl2 were used?

Troyanec [42]

Answer:

6.75 moles of CuCl₂ were used

Explanation:

Given data:

Number of moles of AlCl₃ formed = 4.5 mol

Number of moles of CuCl₂ used = ?

Solution:

Chemical equation:

3CuCl₂ + 2Al → 2AlCl₃ + 3Cu

Now we will compare the moles of CuCl₂ and AlCl₃

AlCl₃ : CuCl₂

2 : 3

4.5 : 3/2×4.5 = 6.75

6.75 moles of CuCl₂ were used.

7 0

3 years ago

Magnesium metal burns in air to form a mixture of magnesium oxide (MgO, M = 40.31) and magnesium nitride (Mg3N2, M = 100.95). A

dedylja [7]

Answer:

26.95 %

Explanation:

Air contains the highest percentage of oxygen and nitrogen gases. Magnesium then combines with both of the gases:

$2 Mg (s) + O_2 (g)\rightarrow 2 MgO (s)$

$3 Mg (s) + N_2 (g)\rightarrow Mg_3N_2 (s)$

Firstly, find the total number of moles of magnesium metal:

$n_{Mg} = \frac{1.000 g}{24.305 g/mol} = 0.041144 mol$

Let's say that x mol react in the first reaction and y mol react in the second reaction. This means:

$x + y = 0.041144 mol$

According to stoichiometry, we form:

$n_{MgO} = x mol, n_{Mg_3N_2} = \frac{y}{3} mol$

Multiplying moles by the molar mass of each substance will yield mass. This means we form a total of:

$m_{MgO} = 40.31x g, m_{Mg_3N_2} = \frac{y}{3} 100.95 g =$

The total mass is given, so we have our second equation to solve:

$40.31x + 33.65y = 1.584$

We have two unknowns and two equations, we may then solve:

$x + y = 0.041144$

$40.31x + 33.65y = 1.584$

Express y from the first equation:

$y = 0.041144 - x$

Substitute into the second equation:

$40.31x + 33.65(0.04144 - x) = 1.584$

$40.31x + 1.39446 - 33.65x = 1.584$

$6.66x = 0.18954$

$x = 0.028459$

$y = 0.041144 - x = 0.012685$

Moles of nitride formed:

$n_{Mg_3N_2} = \frac{y}{3} = 0.0042282 mol$

Convert this to mass:

$m_{Mg_3N_2} = 0.0042282 mol\cdot 100.95 g/mol = 0.4268 g$

Find the percentage:

$\omega_{Mg_3N_2} = \frac{0.4268 g}{1.584 g}\cdot 100\% = 26.95 \%$

7 0

3 years ago