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2 years ago

Which describes the requirements for making an electromagnet?

Chemistry

2 answers:

slava [35]2 years ago

8 0

I think its D(a battery,a length of wire and so on...)

Ivenika [448]2 years ago

5 0

Its D! because a battery and an iron are always connect in a closed parallel circut:)

You might be interested in

A 25.00 mL sample of 0.320 M KOH is titrated with 0.750 M HBr at 25 °C.

Fudgin [204]

Answer:

a. pH = 13.50

b. pH = 13.15

Explanation:

Hello!

In this case, since the undergoing chemical reaction between KOH and HBr is:

$HBr+KOH\rightarrow KBr+H_2O$

As they are both strong. In such a way, since the initial analyte is the 25.00 mL solution of 0.320-M KOH, we first compute the pOH it has, considering that all the KOH is ionized in potassium and hydroxide ions:

$pOH=-log([OH^-])=-log(0.320)=0.50$

Thus, the pH is:

$pH+pOH=14\\pH=14-pOH=14-0.50\\pH=13.50$

Which is the same answer for a and b as they ask the same.

Moreover, once 5.00 mL of the HBr is added, we need to compute the reacting moles of each substance:

$n_{KOH}=0.02500mL*0.320mol/L=0.00800mol\\\\n_{HBr}=0.005L*0.750mol/L=0.00375mol$

It means that since there are more moles of KOH, we need to compute the remaining moles after those 0.00375 moles of acid consume 0.00375 moles of base because they are in a 1:1 mole ratio:

$n_{KOH}^{remaining}=0.00425mol$

Next, we compute the resulting concentration of hydroxide ions (equal to the concentration of KOH) in the final solution of 30.00 mL (25.00 mL + 5.00 mL):

$[OH^-]=[KOH]=\frac{0.00425mol}{0.03000L}=0.142M$

So the pOH and the pH turn out:

$pOH=-log(0.142)=0.849\\pH+pOH=14\\pH=14-pOH=14-0.849\\pH=13.15$

Best regards!

7 0

2 years ago

I WILL GIVE BRAINLY !!!! Modern atomic theory states that the atom is a diffuse cloud surrounding a small, dense nucleus. Identi

Kay [80]

the answer is B because i said so ;)

4 0

3 years ago

The first IE of neon (atomic number 10) is significantly higher than that of argon (atomic number 18) but significantly

lbvjy [14]

Explanation:

Different atoms binds their outermost shell electrons with different amount of energy.

The amount of energy required to remove an electron from an atom is the ionization energy.

Ionization energy measures the readiness of an atom to lose electrons.
From the given problem, we can infer that in group O the ionization energy decreases down the group.
Helium has the highest ionization energy.
Down a group on the periodic table, ionization energy decrease because:

atomic radii increases down the group.
there is an increasing shielding/screening effect of inner shell electrons on the outermost shell electrons.

Learn more:

Ionization energy brainly.com/question/2153804

#learnwithBrainly

8 0

2 years ago

The ph of a 0.30 m solution of a weak acid is 2.67. what is the ka for this acid?

azamat

To determine the Ka of the acid, we can use the equation for the pH of weak acids which is expressed as:

pH = -0.5 log Ka
2.67 = -0.5 log Ka
Ka = 4.571x10^-6

Weak acids are acids that do not dissociate completely in solution. The solution would contain the cations, anions and the acid itself as a compound. Hope this helps.

7 0

3 years ago

Read 2 more answers

1.00 M CaCl2 Density = 1.07 g/mL

Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

$m = 1 mol\times 111 g/mol = 111 g$

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

$M=d\times V=1.07 g/mL\times 1000 mL=1,070 g$

1) The value of %(m/M):

$\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%$

2) The value of %(m/V):

$\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%$

$Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}$

$Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}$

n = Equivalent mass

n = $\frac{\text{molar mass of ion}}{\text{charge on an ion}}$

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 $CaCl_2$ mole has 1 mole of calcium ion)

$n=\frac{40 g/mol}{2}=20$

$=\frac{1 mol}{20 g/mol\times 1L}=0.050 N$

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 $CaCl_2$ mole has 2 mole of chlorine ion)

$n=\frac{35.5 g/mol}{1}=35.5$

$=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N$

Moles of calcium chloride = 1.00 mol

Mass of solvent = Mass of solution - mass of solute

= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

$\frac{1 mol}{0.959 kg}=1.043 mol/kg$

Moles of calcium chloride = $n_1=1mol$

Mass of solvent = 959 g

Moles of water = $n_2=\frac{959 g}{18 g/mol}=53.28 mol$

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842$

7) Mole fraction of water =

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816$

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

$m'=d\times v=1.07 g/ml\times 100 g= 107 g$

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

$11.1\%=\frac{m}{100 mL}\times 100$

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

3 0

2 years ago