90 chemical elements have two letters in their symbols.
Answer:
205.12 atm
Explanation:
Using the ideal gas law equation:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
R = 0.0821 Latm/perK)
T = temperature (K)
n = number of moles (mol)
According to the information in this question;
P = ?
V = 34.25 mL = 34.25 ÷ 1000 = 0.03425L
n = 0.215 mol
T = 125.0°C = 125 + 273 = 398K
Using PV = nRT
P = nRT ÷ V
P = (0.215 × 0.0821 × 398) ÷ (0.03425)
P = 7.025 ÷ 0.03425
P = 205.12 atm
Answer:
MgBr₂ + AgNO₃ => Mg(NO₃)₂ + AgBr
Explanation:
Find the element symbol and charge of each element on the periodic table. For polyatomic ions (nitrate), reference your polyatomic ions chart. Use the "partner's charge" rule to find the number of atoms in each compound.
Charges are written as superscripts. "1" is usually not written, just the + or - sign. The charge of silver is 1, which is the (I) bracket roman numeral 1. It is indicated like that because it is multivalent, meaning it has more than one possible charge.
<u>Write each element as an ion</u> (with the charge).
Magnesium is Mg²⁺
Bromide is Br⁻
Silver(I) is Ag⁺
Nitrate is (NO₃)⁻
<u>Write each compound.</u>
REACTANTS SIDE
Magnesium bromide
Mg²⁺Br⁻ Cross over the partner's charge. Since Br is charge 1, Mg has 1 atom. Since Mg has charge 2, Br has 2 atoms.
MgBr₂
Silver(I) nitrate
Ag⁺(NO₃)⁻
AgNO₃ Both have 1 atom because each partner's charge was 1. You do not need to write brackets if nitrate only has 1 atom.
PRODUCTS SIDE
Magnesium nitrate
Mg²⁺(NO₃)⁻
Mg(NO₃)₂ Nitrate has 2 atoms because magnesium's charge is 2.
Silver(I) bromide
Ag⁺Br⁻
AgBr Both have 1 atom.
Write the compounds into an equation. Reactants go on the left side, products go on the right side. Between the reactants and products, write an arrow.
MgBr₂ + AgNO₃ => Mg(NO₃)₂ + AgBr
Answer:
(a) The slope of distance-time graph gives the speed of the objects.
As the slope of object (B) is the greatest among all, thus B has the largest speed and hence it is the fastest among all the objects.
(b) : As the curve of the objects do not intersect with each other even once, hence they cannot be at the same point on the road at the same time.
(c) : On the y axis (i.e. distance line), 7 small represents 4 km
∴ 1 small line =
7
4
=0.57 km
Position of C at the instant B passes A is 8 km
(d) ; Position of B at the instant it passes C is 2nd small line above 4 km i.e. 4+2×0.57=5.14 km
∴ Distance traveled by B =5.14−0=5.14 kmV
Explanation:
