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Sophie [7]
3 years ago
13

30 divided by 4,800

Mathematics
2 answers:
DedPeter [7]3 years ago
8 0

30/4800 = 3/480

 = 1/160 in fraction form

 = 0.00625 in decimal form

White raven [17]3 years ago
4 0
0.00625 or as a fraction  4800/30<span> </span>
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Which is the coefficient in the expression 7x + 14?
guajiro [1.7K]

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7

Step-by-step explanation:

A coefficient is the number before a variable (x, y, z...etc).

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If sin ∅ = m-n/m+n<br>, find the value of<br>1 + Tan²∅​
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need help.. The midpoint of GH is M(-6,-3). One endpoint is H (-4,4) Find the coordinates of endpoint G.
iren2701 [21]

Answer:

The coordinates of G are: (-8, -10)

Step-by-step explanation:

Mid point formula:

x = \dfrac{x_1+x_2}{2}\\y = \dfrac{y_1+y_2}{2}

Where,

(x,y) are the coordinates of the mid point of the points (x_1,y_1)  and (x_2,y_2).

As per question statement, we are given that:

x_2=-4\\y_2=4

and

x=-6\\y=-3

To find:

x_1 = ?\\y_1 = ?

Solution:

Let us use the mid point formula:

-6 = \dfrac{x_1-4}{2}\\\Rightarrow -12=x_1-4\\\Rightarrow x_1=-12+4\\\Rightarrow \bold{x_1=-8}

-3 = \dfrac{y_1+4}{2}\\\Rightarrow -6=y_1+4\\\Rightarrow y_1=-6-4\\\Rightarrow \bold{y_1=-10}

The coordinates of G are: (-8, -10)

5 0
3 years ago
Calculate the pH of a buffer solution made by mixing 300 mL of 0.2 M acetic acid, CH3COOH, and 200 mL of 0.3 M of its salt sodiu
Lesechka [4]

Answer:

Approximately 4.75.

Step-by-step explanation:

Remark: this approach make use of the fact that in the original solution, the concentration of  \rm CH_3COOH and \rm CH_3COO^{-} are equal.

{\rm CH_3COOH} \rightleftharpoons {\rm CH_3COO^{-}} + {\rm H^{+}}

Since \rm CH_3COONa is a salt soluble in water. Once in water, it would readily ionize to give \rm CH_3COO^{-} and \rm Na^{+} ions.

Assume that the \rm CH_3COOH and \rm CH_3COO^{-} ions in this solution did not disintegrate at all. The solution would contain:

0.3\; \rm L \times 0.2\; \rm mol \cdot L^{-1} = 0.06\; \rm mol of \rm CH_3COOH, and

0.06\; \rm mol of \rm CH_3COO^{-} from 0.2\; \rm L \times 0.3\; \rm mol \cdot L^{-1} = 0.06\; \rm mol of \rm CH_3COONa.

Accordingly, the concentration of \rm CH_3COOH and \rm CH_3COO^{-} would be:

\begin{aligned} & c({\rm CH_3COOH}) \\ &= \frac{n({\rm CH_3COOH})}{V} \\ &= \frac{0.06\; \rm mol}{0.5\; \rm L} = 0.12\; \rm mol \cdot L^{-1} \end{aligned}.

\begin{aligned} & c({\rm CH_3COO^{-}}) \\ &= \frac{n({\rm CH_3COO^{-}})}{V} \\ &= \frac{0.06\; \rm mol}{0.5\; \rm L} = 0.12\; \rm mol \cdot L^{-1} \end{aligned}.

In other words, in this buffer solution, the initial concentration of the weak acid \rm CH_3COOH is the same as that of its conjugate base, \rm CH_3COO^{-}.

Hence, once in equilibrium, the \rm pH of this buffer solution would be the same as the {\rm pK}_{a} of \rm CH_3COOH.

Calculate the {\rm pK}_{a} of \rm CH_3COOH from its {\rm K}_{a}:

\begin{aligned} & {\rm pH}(\text{solution}) \\ &= {\rm pK}_{a} \\ &= -\log_{10}({\rm K}_{a}) \\ &= -\log_{10} (1.76 \times 10^{-5}) \\ &\approx 4.75\end{aligned}.

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3 years ago
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Kruka [31]

Answer: 27

Step-by-step explanation:

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2 years ago
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