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2 years ago

Help !!I can't understand it

Mathematics

2 answers:

siniylev [52]2 years ago

8 0

a.) The color of Aqueous solution turns light green from blue, due to formation of iron sulphate .

b.) The color of the Aqueous solution turns colorless from blue due to formation of zinc sulphate.

2. Bells used in temples aren't made up of wood because wood isn't sonorous and hence can't produce ringing sound, and hence metals are used in making bells.

3. phosphorous is kept in water because it is highly reactive and reacts vigorously with air if left free in atmosphere .

a.) CuSO4 + Mg =》MgSO4 + Cu

this is a single displacement reaction, where Magnesium being more reactive than Copper, displaces copper from its compound (copper sulphate) to form magnesium sulphate.

b. H2SO4 + Zn =》ZnSO4 + H2

this is a single displacement reaction,

where Zinc being more reactive than

hydrogen, displaces hydrogen from its

compound (sulphuric acid) to form Zinc sulphate.

Helen [10]2 years ago

7 0

a.) The color of Aqueous solution turns

The color of Aqueous solution turnslight green from blue, due to formation

The color of Aqueous solution turnslight green from blue, due to formationof iron sulphate.

b.) The color of the Aqueous solution

The color of the Aqueous solutionturns colorless from blue due to

The color of the Aqueous solutionturns colorless from blue due toformation of zinc sulphate.

Bells used in temples aren't made upof wood because wood isn't sonorousand hence can't produce ringing sound,and hence metals are used in makingbells.

phosphorous is kept in water becauseit is highly reactive and reacts vigorouslywith air if left free in atmosphere

a.) CuSO4+ Mg MgS04 +Cu

this is a single displacement reaction,where Magnesium being more reactivethan Copper, displaces copper from itscompound (copper sulphate) to formmagnesium sulphate.

b. H2S04 +Zn =) ZnS04+H2

b. H2S04 +Zn =) ZnS04+H2this is a single displacement reaction,where Zinc being more reactive thanhydrogen, displaces hydrogen from itscompound (sulphuric acid) to form Zincsulphate.

Step-by-step explanation:

I hope it helps

have a great day

#Liliflim

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The square pyramid pictured below has a surface area of

Evgen [1.6K]

Answer:

Step-by-step explanation:

Triangles

Area of a triangle = 1/2 b * slanted height

b = 6 m

h = 9 m

Area of 1 triangle = 1/2 * 6 * 9

Area of 1 triangle = 27

Area of all 4 triangles = 4*27 = 108 m^2

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

Base area

The base is a square. All 4 sides are equal.

The formula for a square is s^2

Area = (6 m)^2

Area = 36 m^2

Total area

Total Area = 108 m^2 + 36 m ^2

Total Area = 144 m^2

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3 years ago

Read 2 more answers

If f(x) = 2x2 - 5 and g(x) = 3x + 3, find (f - g)(x).

Vera_Pavlovna [14]

Answer: A. 2x2 - 3x - 8

Just subtract the functions and combine like terms
(f - g)(x) = f(x) - g(x)
(f - g)(x) = 2x^2 - 5 -(3x + 3)
(f - g)(x) = 2x^2 - 5 - 3x - 3
(f - g)(x) = 2x^2 - 3x - 8

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3 years ago

I need help with this question

schepotkina [342]

V=763.02
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3 years ago

Linda paid $48 for a jacket that was on sale for 25% of the original price. What was the original price of the jacket

kap26 [50]

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3 0

3 years ago

Determine any asymptotes (Horizontal, vertical or oblique). Find holes, intercepts and state it's domain.

vesna_86 [32]

Factorize the denominator:

$\dfrac{x^2-4}{x^3+x^2-4x-4}=\dfrac{x^2-4}{x^2(x+1)-4(x+1)}=\dfrac{x^2-4}{(x^2-4)(x+1)}$

If $x\neq\pm2$ , we can cancel the factors of $x^2-4$ , which makes $x=-2$ and $x=2$ removable discontinuities that appear as holes in the plot of $g(x)$ .

We're then left with

$\dfrac1{x+1}$

which is undefined when $x=-1$ , so this is the site of a vertical asymptote.

As $x$ gets arbitrarily large in magnitude, we find

$\displaystyle\lim_{x\to-\infty}g(x)=\lim_{x\to+\infty}g(x)=0$

since the degree of the denominator (3) is greater than the degree of the numerator (2). So $y=0$ is a horizontal asymptote.

Intercepts occur where $g(x)=0$ ( $x$ -intercepts) and the value of $g(x)$ when $x=0$ ( $y$ -intercept). There are no $x$ -intercepts because $\dfrac1{x+1}$ is never 0. On the other hand,

$g(0)=\dfrac{0-4}{0+0-0-4}=1$

so there is one $y$ -intercept at (0, 1).

The domain of $g(x)$ is the set of values that $x$ can take on for which $g(x)$ exists. We've already shown that $x$ can't be -2, 2, or -1, so the domain is the set

$\{x\in\mathbb R\mid x\neq-2,x\neq-1,x\neq2\}$

8 0

3 years ago

Help !!I can't understand it ​

Help !!I can't understand it