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3 years ago

5

How many base units exist?

1 answer:

Oksi-84 [34.3K]3 years ago

4 0

Answer:

7

Explanation:

second

metre

kilogram

ampere

kelvin

mole

candela

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What is the percent yield of ferrous sulfide if the actual yield is 220.0 g and the theoretical yield is 275.6 g

Anika [276]

To get the percent yield, we will use this formula: ((Actual Yield)/(Theoretical Yield)) * 100% Values given: actual yield is 220.0 g theoretical yield is 275.6 g Now, let us substitute the values given. (220.0 grams)/(275.6 grams) = 0.7983 Then, to get the percentage, multiply the quotient by 100. 0.7983 (100) = 79.83% Among the choices, the most plausible answer is 79.8% <span>
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4 0

3 years ago

Read 2 more answers

ziro4ka [17]

<h2>Answer:0.5 moles,1 mole</h2>

Explanation:

$Sr(NO_{3})_{2}$ is readily soluble in water.

$Sr(NO_{3})_{2}$ dissolves in water to give $Sr^{2+}$ ions and $NO_{3}^{-}$ ions.

The reaction is

$Sr(NO_{3})_{2}$ → $Sr^{2+}$ $+2NO_{3}^{-}$

So, $1$ mole of $Sr(NO_{3})_{2}$ gives $1$ mole of $Sr^{2+}$ and $2$ moles of $NO_{3}^{-}$ .

So, $0.5$ moles of $Sr(NO_{3})_{2}$ gives $0.5$ moles of $Sr^{2+}$ and $1$ mole of $NO_{3}^{-}$ .

5 0

3 years ago

What is the percent composition of Br in CuBr3?

tekilochka [14]

Answer:

about 79% (79.04369332 to be exact)

Explanation:

Percent composition=(Molar mass of element x amount of it)/Molar mass of compound x 100

Br= 3 x 79.9/303.25 x100=79.04369332

6 0

3 years ago

Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, th

Gre4nikov [31]

Answer : The equilibrium concentration of NO is, 0.0092 M.

Solution :

First we have to calculate the concentration of NO.

$\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M$

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

Initially conc. 0 0 0.1576

At eqm. (x) (x) (0.1576-2x)

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$0.0153=\frac{(0.1576-2x)^2}{(x)\times (x)}$

By solving the term, we get:

$x=0.0742,0.0839$

Neglecting the 0.0839 value of x because it can not be more than initial value.

Thus, the value of 'x' will be, 0.0742 M

Now we have to calculate the equilibrium concentration of NO.

Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M

Therefore, the equilibrium concentration of NO is, 0.0092 M.

5 0

3 years ago

In each of the following blanks, only enter a numerical value.

tatiyna

Answer:

1) 1,... 2

2) 18

3) n= 3 and I=1

Explanation:

1) when l= 0, its an s-sub-level, and only 1 orbital is possible which can carry only 2-electrons

2) the maximum number of electron is given by 2n^2= 2×3^2= 18

3) in 3p, the coefficient of p is the value of n= 3 and l-value of P is 1

5 0

3 years ago