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4 years ago

Name four ways that bacteria are helpful to people

1 answer:

8 0

<span>gives you dairy products
helps fight disease
helps digest food
They are used in the manufacturing of antibiotics.</span>

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What is the definition of thermal expansion

MissTica

Answer:

Thermal expansion is the tendency of matter to change its shape, area, and volume in response to a change in temperature.

Explanation:

8 0

4 years ago

What element is undergoing oxidation in the following reaction?. . CH4(g) +2O2 (g) -> CO2(g) +2H2O(g). . . option . a) C. b)

SVETLANKA909090 [29]

The correct answer is a.) C. Oxidation occurs when an element loses electrons and increases its oxidation state. In the chemical equation, Carbon in CH4 has an oxidation state of 4- while Carbon in CO2 has an oxidation of 4+.

4 0

3 years ago

Read 2 more answers

How many moles of He are in 16 g of the element? 1.0 mol 4.0 mol 8.0 mol 16 mol

Lady bird [3.3K]

Answer: 4 Mole

Explanation:

16/4 = 4

4 0

4 years ago

If a solution containing 51.429 g of mercury(ii) perchlorate is allowed to react completely with a solution containing 16.642 g

OLga [1]

Hg(No3)2 +NaSO4 --->2NaNO3 + HgSO4(s)
calculate the moles of each reactant
moles=mass/molar mass

moles of Hg(NO3)2= 51.429g/ 324.6 g/mol(molar mass of Hg(NO3)2)=0.158 moles

moles Na2SO4 16.642g/142g/mol= 0.117 moles of Na2SO4

Na2SO4 is the limiting reagent in the equation and by use mole ratio Na2So4 to HgSO4 is 1:1 therefore the moles of HgSO4 =0.117 moles

mass of HgSO4=moles x molar mass of HgSo4= 0.117 g x 303.6g/mol= 35.5212 grams

7 0

4 years ago

Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures ar

kakasveta [241]

Answer:

The answer is " $\bold{0.525\ \ atm^{-1}}$ "

Explanation:

Given equation:

$2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)$

Given value:

$\Delta rH =-198.2 \ \ \frac{KJ}{mol}$

$Kp=1100 \ K$

$\Delta x = 2-(2+1)\\\\$

$= 2-(2+1)\\\\= 2-(3)\\\\= -1$

$\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm) &1-2x&0.5-x&2x\end{array}\right$

calculating the total pressure on equilibrium= $(1-2x)+(0.5-x)+2x \ atm\\\\$

$= 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\$

$\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to x= 0.15\\\\$

calculating the pressure in $So_2$ :

$= (1-2 \times 0.15)$

$= 1-0.30 \\\\ =0.70 \ atm$

calculating the pressure in $O_2$ :

$= (0.5- 0.15)\\\\= 0.35 \ atm \\$

calculating the pressure in $So_3$ :

$= (2 \times 0.15)\\\\= (.30) \ atm \\\\$

Calculating the Kp at 1100 K:

$= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\= 0.5247 \ \ or \ \ 0.525 \ \ atm^{-1} \\\\$

4 0

4 years ago