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3 years ago

Need help I don't remember what to do?

Chemistry

1 answer:

Lina20 [59]3 years ago

4 0

I'm just doing the ones that you don't have numbers already for.
2.) just leave it alone and it's correct
3.) Mg + 2AgNo3 --> Mg(No3)2 + 2Ag
5.) just leave it alone and it's correct
8.) 10C3H8O3 + 15O2 --> 30CO2 + 4H2O
10.) P4 + 6Br2 --> 4PBr3
12.) 2FeCl3 + 6NaOH --> 2Fe(OH)3 + 6NaCl
13.) 2CH3OH + 3O2 --> 2CO2 + 4H2O
14.) 2Al + 3Cu(NO3)2 --> 2Al(NO3)3 + 3Cu
15.) 3CaCl2 + 2K3AsO4 --> Ca3(AsO4)2 + 6KCl
16.) 2NH3 --> N2 + 3H2
17.) 2H3PO4 + 3Ba(OH)2 --> Ba3(PO4)2 + 6H2O
19.) Mg3N2 + 6H2O --> 3Mg(OH)2 + 2NH3
I hope this helps you!!

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Why does a mechanical wave require a medium for transmission?

Ganezh [65]

The medium provides an opposing force to slow down the wave.

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3 years ago

Consider the reaction between iron (III) oxide, Fe2O3 and carbon monoxide, CO. Fe2O3 + 3CO --> 2Fe + 3CO2 In the process, 213

s344n2d4d5 [400]

To calculate the mass of Fe formed in a) we get first the limiting reactant between Fe2O3 and CO. Given the masses, the ratio of Fe2O3 is 1.33 while that of CO is 1.67. Hence the limiting reagent is Fe2O3. The mass of Fe formed is 148.98 grams. In b, the needed CO is only 112.04 grams. Hence, the excess is 27. 96 grams.

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3 years ago

Balance the following redox equation and identify the element oxidized, the element reduced, the oxidizing agent, and the reduci

earnstyle [38]

The charge of Br changed from –1 to 0, therefore it is the element which is oxidized. Since it is oxidized then Br is also the reducing agent.

The charge of Mn changed from +4 to +2 therefore it is the element which is reduced. Since Mn is reduced, then MnO2 is the oxidizing agent.

The half –reactions are:

Br: 2Br --> Br2 + 2e-

Mn: MnO2 --> Mn2+

First balance oxygen by adding H2O:

MnO2 --> Mn2+ + 2H2O

Then balance hydrogen by adding H+ ions:

4H+ + MnO2 --> Mn2 + 2H2O

Then the appropriate electrons:

4e- + 4H+ + MnO2 --> Mn2 + 2H2O

Multiply the half-reaction of Br by 2 because the half-reaction of Mn has 4 electrons.

4Br --> 2Br2 + 4e-

Combine the two half reactions and cancel common factors:

4Br- + 4H+ + MnO2 --> 2Br2 + Mn2 + 2H2O

6 0

2 years ago

Fill in the missing data point. Show all calculations leading to an answer.

Aleksandr-060686 [28]

Answer:

1090 mmHg

Explanation:

We know that with gases we must use a Kelvin temperatures, so let’s try a plot of pressure against the Kelvin temperature.

We can create a table as follows

t/°C T/K p/mmHg

10 283 726

20 293 750

40 313 800

70 343 880

100 373 960

150 423 ???

I plotted the data and got the graph in the figure below.

It appears that pressure is a linear function of the Kelvin temperature.

y = mx + b

where x is the slope and b is the y-intercept.

===============

Calculate the slope

I will use the points (275, 700) and (380, 975).

Slope = Δy/Δx = (y₂ - y₁)/(x₂ -x₁) = (975 -700)/(380 – 275) = 275/105 = 2.619

So,

y = 2.619x + b

===============

Calculate the intercept

When x = 275, y = 700.

700 = 2.619 × 275 + b

700 = 720 + b Subtract 720 from each side and transpose.

b = -20

So, the equation of the graph is

y = 2.619x -20

===============

Calculate the pressure at 423 K (150°C)

y = 2.619 × 423 - 20

y = 1110 - 20

y = 1090

At 150 °C, the pressure 1090 mmHg.

The point is approximately at the position of the black dot in the graph.

7 0

3 years ago

There are three naturally occurring isotopes of the hypothetical element hilarium 45Hi, 46Hi, and 48Hi. The percentages of these

den301095 [7]

Answer:

46.761g/mol

Explanation:

Given parameters:

Element = Hilarium , Hi

Isotopes: Hi- 45, Hi-46 and Hi- 48

Natural abundance of Hi-45 = 18.3%

Hi-46 = 34.5%

Hi-48 = 47.2%

Unknown:

Atomic weight of naturally occurring Hilarium = ?

Solution:

Isotopes have been studied extensively by mass spectrometry. The method is used to determine the proportion/percentage/fraction by which each of the isotopes of an element occurs in nature. The proportion is called geonormal abundance. From this we can calculate the atomic weight of an element.

We can use the expression below to find this value:

Atomic weight = m₄₅α₄₅ + m₄₆α₄₆ + m₄₈α₄₈

m is the atomic mass of each isotope and α is the abundance

Atomic weight = (45 x $\frac{18.3}{100}$ ) + (46 x $\frac{34.5}{100}$ ) + (48 x $\frac{47.2}{100}$ )

Atomic weight of Hi = 8.235 + 15.870 + 22.656 = 46.761g/mol

6 0

3 years ago